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This lab book is a synopsis of the setup, required apparatus, and data collection and analysisfor each of our Modern Physics Laboratory experiments.
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Advanced Experiments in Physics. Physics 306
University of Wisconsin-Parkside
Jeffrey R. Schmidt
October 1997, 1999, 2001, 2003, Revision: December 2005
NaI
Photocathode
Dynodes
1
Contents
1 Introduction 4
2 Nuclear Counting Statistics 52.1 Apparatus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 Geiger tube calibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.4 Experiment I; Normal distributions . . . . . . . . . . . . . . . . . . . . . . . 82.5 Experiment II. Hypothesis testing and goodness of fit . . . . . . . . . . . . . 14
3 The Half-life of Ba137 203.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.2 Apparatus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.3 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.4 Data and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.5 Background material; the Nuclear Shell Model . . . . . . . . . . . . . . . . . 27
4 Determination of the Speed of Light 36
5 Determination of Cp
Cvfor a gas 40
5.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405.2 Apparatus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425.3 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425.4 Data and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.5 Molecular spectra and the equipartition theorem . . . . . . . . . . . . . . . . 455.6 Safe Handling of High Pressure Gas Cylinders . . . . . . . . . . . . . . . . . 495.7 Lissojous program . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
6 Electron Diffraction and Bragg Scattering 536.1 Apparatus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546.3 Data and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 576.4 Crystal structure determination by scattering . . . . . . . . . . . . . . . . . 616.5 Simulation of scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
7 The Photoelectric Effect 697.1 Apparatus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 697.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707.3 Procedure and data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
8 Measuring Planck’s constant (solid state version) 738.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 738.2 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
8.2.1 Finding roots by bisection . . . . . . . . . . . . . . . . . . . . . . . . 798.2.2 Newton’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
2
9 Helium Ionization Potential; The Franck-Hertz Experiment 839.1 The old version of the experiment; apparatus . . . . . . . . . . . . . . . . . . 849.2 Data and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 869.3 Second version . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
10 The Rydberg Constant 8910.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8910.2 Atomic Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9410.3 Apparatus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9810.4 Data and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10010.5 Diffraction gratings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
11 Measurement of unknown spectral lines 10611.1 The Mercury data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10711.2 First order calibration curve . . . . . . . . . . . . . . . . . . . . . . . . . . . 10811.3 Measuring Sodium lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
12 Stefan-Boltzmann Radiation Law 10912.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10912.2 Luminosity or radiance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11512.3 Apparatus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11612.4 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12212.5 Data and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
13 Measurement of the Compton Edge 12613.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12613.2 Apparatus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12913.3 Scintillation counters and photomultipliers . . . . . . . . . . . . . . . . . . . 13013.4 Stage 1; calibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13413.5 Stage II; photopeak and edge measurements . . . . . . . . . . . . . . . . . . 135
14 Energy versus Momentum for Relativistic Electrons 13514.1 Apparatus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13614.2 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13614.3 Data and analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
15 Michelson-Interferometer and the Index of Refraction of Air 138
16 Angular distribution of emitted radiation 14216.1 Polarization and angular distributions . . . . . . . . . . . . . . . . . . . . . . 14516.2 Angular correlations of γ emissions . . . . . . . . . . . . . . . . . . . . . . . 14816.3 A simulated experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
3
17 A simulated experiment; Geiger-Marsden experiment 15417.1 The numerical experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16317.2 Raw data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16417.3 Problem 1. Analysis of simulated lab data . . . . . . . . . . . . . . . . . . . 16517.4 The Thompson model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16717.5 Monte Carlo scattering simulation . . . . . . . . . . . . . . . . . . . . . . . . 16817.6 Energy dependence in the Geiger-Marsden experiment . . . . . . . . . . . . 17017.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17417.8 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
18 Appendix 17818.1 Appendix I. Reformatting Lab Data with Sed and Awk . . . . . . . . . . . . 17818.2 Appendix II; Nonlinear Least Squares error matrix method . . . . . . . . . . 18418.3 Appendix III. The Plotting of Lab Data with GNU plotutils . . . . . . . . . 18718.4 Appendix IV. Preparing a lab report with LaTeX . . . . . . . . . . . . . . . 19118.5 Appendix V. The Oscilloscope . . . . . . . . . . . . . . . . . . . . . . . . . . 19518.6 Appendix VI. Support software . . . . . . . . . . . . . . . . . . . . . . . . . 196
1 Introduction
This lab book is a synopsis of the setup, required apparatus, and data collection and analy-sis for each of our Modern Physics Laboratory experiments. The data for each experimentare real, taken in 1996, 1997, 1999, and 2001. The book can serve as an outline and basicreference for the setup of the experiments and their proper analysis.I do not generally teach this course at the present time, so this has been prepared as acourtesy to the students and any person who must teach the course for the first time, butmostly in preparation for the day when I will be teaching it more regularly.
In addition to twelve basic experiments, and a variation on one or two of these, there are anumber of simulated experiments; things that would be interesting to do if we had the equip-ment. These include the Geiger-Mardsen experiment, Laue scattering, and γ − γ angularcorrelations. There are two numerical labs in the Geiger-Marsden section, one verifying theenergy dependence of the scattering cross section, another verifying the Rutherford formulafor the number of α particles scattered from a gold foil. Hopefully I will add more simulationsas time allows.
c©1997, 2003, 2005 Jeffrey R. Schmidt
The experiments
4
2 Nuclear Counting Statistics
Nuclear decays are random processes that provide an excellent context in which to studybasic statistics and distribution functions. We will determine whether or not nuclear decaysare random events by counting radioactive decays with a Geiger tube and a scalar-counter.
Freq Volts Freq Time Start Count Time Reset Power
Geiger Tube Voltage
10 10hr
100 1hr
1k 10min
10k 1min
100k 10s
10M 1s
Count TimePresetSARGENT−WELCH | SCALAR \ TIMER
PresetMode
Ext 50/60 Hz
2.1 Apparatus
1. Geiger tube. Welch Scientific #1216.2. Scaler-timer. Sargent-Welch model S-72095-10.3. Radioactive source. 0.1µc Cs137, Amerschorn-Searle 184471.
2.2 Background
In at least four experiments in 306 you will study radiations emitted from unstable isotopes.Some of these are given in the table below
Isotope Emission Half-life
55Cs137, 56Ba
137 β−; 0.514MeV 27 yearγ; 0.662MeV
27Co60, 58Ni
60 β−; 0.312MeV 5.2 yearγ; 1.172MeVγ; 1.333MeV
58Ce144, P r144 β−; 0.320MeV 285 days
β; 0.184MeVγ; 0.134MeVγ; 0.081MeV
11Na22 β+; 0.542MeV 2.6 year
γ; 1.277MeVγγ; 0.511MeV(From β−β+ → γγ)
5
The Geiger tube is used to detect the decay particles emitted by the Cesium sample. Thescalar-timer counts the number of decays from the source in ten second intervals. This is thedata that will be statistically analyzed. A selection of our Geiger counters is shown below.The counter is a cylindrical glass tubewith a thin tungsten wire running down the center,and a coaxial metal cylinder enclosed in the glass. The tube is filled with air at a very lowpressure, a few torr. A voltage typically around 1000V is applied between the wire andcylinder, with wire positive and cylinder negative. This voltage is slightly below dielectricbreakdown voltage for the gas in the tube.
R
Tungsten wirecylinder
1000 V
Ionizing radiation entering the tube will free ions from air molecules, which will in turn beaccelerated by the potential difference in the tube. This causes more collisions, liberatingmore ions and creating a current flow from wire to cylinder and through the external resistorR. The back-voltage across R combines with the voltage on the tube and diminishes it,causing the net potential difference between wire and cylinder to drop below the criticalvalue needed to sustain the cascade of ions. This shuts off the current, but not before it canbe registered on the supporting electronics connected to the tube.Proper operation of the tube depends critically on the voltage applied to it. Too high a volt-age and the tube discharges spuriously from dielectric breakdown, registering false countseven with no ionizing radiation present. To find the optimal operating voltage, the tubemust be calibrated by counting a constant source of ionizing radiation and generating thecharacteristic curve of counts versus voltage for the particular tube.
2.3 Geiger tube calibration
A constant source is placed about 20 cm from the tube, and is counted at a variety of ap-plied voltages. For voltages below the threshold value there will be no counts detected bythe tube, as the voltage is increased above this point. the number of registered counts willincrease and reach a plateau. The center of this plateau is the desired operating voltage.As the voltage continues to climb, the number of counts blows up as the tube registers spu-rious counts from dielectric breakdowns.
6
Sample calibration data This data was taken by two students in 1995 for one of theSargent-Welch Geiger-Muller tubes used in the Modern Physics lab course.
Voltage (volts) Counts Voltage (volts) Counts
507 0 954 1192550 0 1003 1238603 0 1058 1258653 0 1099 1206705 0 1162 1221750 0 1202 1256774 158 1263 1295805 886 1309 1371853 1120 1354 1507898 1185 1399 1712
the plateau is fairly evident from the graph
Geiger Plateau
400 600 800 1000 1200 14000
500
1000
1500
2000
Voltage V
Cou
nts
7
which indicates an operating voltage around 1038 volts.
In the experiment we will attempt to verify that the frequency with which the Geiger countergets X counts per time T is distributed as a random variable, demonstrating that nucleardecays are random.If this hypothesis is true, in time dt there is probability p that a nucleus in the sample willdecay, (1− p) that it will not, so in time T = N dt we should have a probability of gettingM counts of
℘(M) =
(
N
M
)
pM(1− p)N−M → µM
M !e−µ
where µ = Np, in other words a Poissonian distribution.
2.4 Experiment I; Normal distributions
We can use nuclear decays to learn a little bit about the Normal distribution, assuming thatthe decay process is random and Normally distributed.For data that is distributed Normally, meaning by a frequency determined by a PDF equalto that of a Normal distribution, how many data should occur within one standard deviationof the mean?
∫ σ
0
1√2πσ2
e−x2
2σ2 dx = 0.34135
∫ 2σ
0
1√2πσ2
e−x2
2σ2 dx = 0.47725
∫ 3σ
0
1√2πσ2
e−x2
2σ2 dx = 0.4985
In other words, within one standard deviation of the mean you will find 68.26% of all of thedata (both sides of the mean), and within 2σ of the mean you will find 95.45% of the data.Consider the following set of 104 Geiger counter counts for a Barium sample, each takenover the same time period.
8
459 422 434 410 434 438 454 421 390 422 438 436 444449 398 418 407 399 445 424 440 430 391 421 433 401430 449 437 421 460 414 447 396 394 430 418 392 414387 405 409 406 449 437 433 444 396 394 389 426 453410 415 363 438 422 377 410 408 406 431 438 434 413396 436 461 402 377 443 411 383 433 420 393 424 433458 459 444 381 389 438 435 374 440 409 431 433 455427 407 435 405 442 423 398 423 417 423 428 423 406
We can write a little C program to read this data (as a long list of numbers, one per line)from a file into an array, from which an unbiased estimator for the mean and standard de-viation can be gotten, as well as a largest and smallest value.
/* simple data sorting, mean and standard deviation program */
/* gcc sort.c -lm */
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int numbers[200];
double mean,sigma;
int largest,smallest,n,i,sum;
FILE *fptr;
main(int argc, char *argv[])
if(argc!=2)
printf("./sort datafile\n");
exit(0);
fptr=fopen(argv[1],"r"); /* read data file named on command line */
n=0;
do
fscanf(fptr,"%d", &numbers[n]);
n=n+1;
while(!feof(fptr));
n=n-1; /* we overshot */
/* get largest, smallest, do stats */
smallest=numbers[0];
largest=numbers[0];
for(i=1;i<n;i++)
if(numbers[i] < smallest) smallest=numbers[i];
if(numbers[i] > largest) largest=numbers[i];
mean=0.0;
9
for(i=0;i<n;i++)
mean=mean+(double)numbers[i];
mean=mean/(double)n;
sigma=0.0;
for(i=0;i<n;i++)
sigma=sigma+((double)numbers[i]-mean)*((double)numbers[i]-mean);
sigma=sqrt(sigma/(double)n);
/* print the number of data, smallest, largest, mean and standard deviation */
printf("%d\t%d\t%d\t%f\t%f\n",n,smallest,largest,mean,sigma);
/* how many are within sigma of mean, to left ? */
sum=0;
for(i=0;i<n;i++)
if(numbers[i]<= mean && numbers[i]>= mean-sigma) sum=sum+1;
printf("There are %d between mean=%f and mean-sigma=%f\n",sum,
mean,mean-sigma);
/* how many are within sigma of mean, to right ? */
sum=0;
for(i=0;i<n;i++)
if(numbers[i]>= mean && numbers[i]<= mean+sigma) sum=sum+1;
printf("There are %d between mean=%f and mean+sigma=%f\n",sum,
mean,mean+sigma);
fclose(fptr);
This program can be downloaded from the 499 archive. If we run it on our data we obtain
gcc -o sort sort.c -lm
./sort data2
104 363 461 420.846154 21.848591
There are 25 between mean=420.846154 and mean-sigma=398.997562
There are 42 between mean=420.846154 and mean+sigma=442.694745
Notice that there are 25 + 42 = 67 of the 104 data within one standard deviation of themean, confirming the notion that the number of nuclear decays per fixed time interval is mostlikely distributed Normally; this is 64.4%, agreeing quite well with the values computed fora Normal distribution.
If we want to sort these data to create a frequency plot, graphing the number of data that fallwithin a certain lower and upper limit, suitable for graphing, we use another short program.
10
/* simple data sorting, to create histogram plot */
/* gcc pidgeon.c -lm */
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int numbers[200],bins[100];
FILE *fptr;
main(int argc, char *argv[])
float div;
int largest,smallest,n,N,i,which,sum;
if(argc!=5)
printf("./pidgeon datafile min max number\n");
exit(0);
fptr=fopen(argv[1],"r"); /* read data file named on command line */
smallest=atof(argv[2]);
largest=atof(argv[3]);
N=atoi(argv[4]);
div=((float)largest-(float)smallest)/(float)N;
n=0;
do
fscanf(fptr,"%d", &numbers[n]);
n=n+1;
while(!feof(fptr));
n=n-1; /* we overshot */
/* zero all bins */
for(i=0;i<N;i++)
bins[i]=0;
/* pidgeon hole each bit of data */
for(i=0;i<n;i++)
which=(int)floor(((float)numbers[i]-(float)smallest)/div);
bins[which]=bins[which]+1;
for(i=0;i<N;i++)
printf("%f\t%d\n",smallest+(float)i*div+div/2.0,bins[i]);
This we can pass parameters to. For example, our data above could be made into a histogramwith 22 divisions between 360 and 470 with
gcc -o pidgeon pidgeon.c -lm
11
./pidgeon data2 360 470 11
365.000000 1
375.000000 3
385.000000 5
395.000000 12
405.000000 12
415.000000 11
425.000000 16
435.000000 24
445.000000 12
455.000000 6
465.000000 2
or piped directly into a graph with
./pidgeon data2 360 470 11 | graph -T X
This program is giving the midpoint of each division, versus the population of that division.
You may wish to determine exactly what fraction of the total data falls between xmin andxmax for a Normal distribution of variance σ2. For zero mean this is the total number ofdata points times the probability
℘ =∫ xmax
xmin
1√2πσ2
e−x2
2σ2 dx
which is computed with the program below.
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define PI 3.1415926
double dx,x,sum;
double xmin,xmax,sigma;
int n;
main(int argc, char *argv[])
if(argc != 4)
printf("./normal xmin xmax sigma\n");
exit(0);
xmin=(double)atof(argv[1]);
xmax=(double)atof(argv[2]);
sigma=(double)atof(argv[3]);
12
dx=(xmax-xmin)/10000.0;
sum=0.0;
for(n=0;n<10000;n++)
x=xmin+(double)n*dx;
sum=sum+exp(-x*x/(2.0*sigma*sigma));
sum=sum*dx/(sigma*sqrt(2.0*PI));
printf("%f\n",sum);
You can easily shift the mean of the distribution to the point of your choice. You might wantto use this program for the next nuclear counting experiment, to fit your count frequency toa Gaussian Normal distribution rather than the Poissonian example.
You may wish to simulate the experiment before you actually perform it. The C programbelow will generate N Normally distributed random numbers with mean mean and standarddeviation sigma by the accept-reject algorithm. You can use it to simulate the outcomesof he nuclear counting experiments, and use the other programs to analyze the resulting data.
/* gcc -o accept_reject accept_reject.c -lm */
/* creates Normal deviates */
#include <stdlib.h>
#include <math.h>
#include <stdlib.h>
#define PI 3.1415926
int n;
float f( float x, float mean, float sigma);
main(int argc, char *argv[])
float m1,m2,test;
float min,max,sigma,mean;
int N, crap;
if(argc != 4)
printf("./random mean sigma number\n");
exit(0);
mean=(float)atof(argv[1]);
sigma=(float)atof(argv[2]);
N=atoi(argv[3]); /* how many to generate */
srand(17); /* initialize random number generator */
min=mean-5.0*sigma;
max=mean+5.0*sigma;
n=0;
13
do
/* get a random number between mean-4*sigma, mean+4*sigma */
m1=min+(max-min)*(float)rand()/(float)RAND_MAX;
m2=f(m1,mean,sigma);
test=((1.2533141/sigma))*(float)rand()/(float)RAND_MAX;
if(m2>test)
crap=floor(m1);
printf("%d\n", crap);
n=n+1;
while(n<N);
float f( float x, float mean, float sigma)
/* Normal distribution */
float y;
y=(x-mean)/sigma;
return( ((1.2533141/sigma))*exp(-y*y/2));
2.5 Experiment II. Hypothesis testing and goodness of fit
Below is a table of the number of counts for 600 five second intervals. This data was takenin 1997.The Geiger tube voltage has been set at 1040 volts, the scaler-timer counts for 5 seconds ata time with 5 second intervals between counting periods, so that we have time to record ournumbers. The more counts you make, the better will be your results. The number400 is a minimum, I suggest that you take twice that many. This requires counting for overtwo hours. Do it in 30 minute shifts with your lab partners.
14
459 422 434 410 434 438 454 421 390 422 438 436 444 422 427449 398 418 407 399 445 424 440 430 391 421 433 401 425 424430 449 437 421 460 414 447 396 394 430 418 392 414 423 432387 405 409 406 449 437 433 444 396 394 389 426 453 423 422410 415 363 438 422 377 410 408 406 431 438 434 413 432 429396 436 461 402 377 443 411 383 433 420 393 424 433 390 450458 459 444 381 389 438 435 374 440 409 431 433 455 455 392427 407 435 405 442 423 398 423 417 423 428 423 406 421 424418 453 418 398 408 438 443 412 433 440 417 425 427 403 414398 401 408 401 408 454 390 437 410 416 408 446 409 425 396388 423 441 394 414 420 424 396 417 405 429 395 381 438 403445 393 424 431 419 404 398 423 410 445 430 402 428 407 453429 405 389 482 369 406 417 426 402 393 456 428 402 407 406435 433 435 417 450 426 445 443 446 424 393 433 438 424 405409 428 410 400 408 412 428 435 431 395 419 431 430 428 461434 433 406 404 403 404 420 425 417 433 392 391 439 431 419417 405 398 435 407 446 437 458 402 415 432 408 383 374 418400 387 410 360 385 445 427 439 423 465 434 413 435 416 430442 437 418 447 456 426 413 396 435 430 433 360 434 428 454396 408 413 406 406 391 431 437 450 382 413 422 406 437 393438 387 430 425 415 429 430 393 381 437 479 409 396 412 428426 423 362 424 411 432 399 432 397 422 424 410 449 415 478420 416 421 417 417 408 466 436 451 412 433 407 419 460 394427 415 432 429 410 449 421 381 423 435 402 419 432 453 416447 404 386 434 445 399 393 461 399 417 434 432 425 429 445450 402 412 407 414 423 420 404 405 425 385 448 452 435 418419 436 425 475 418 426 378 469 429 438 453 432 418 368 400419 439 423 356 428 382 415 422 402 437 444 417 364 448 443383 435 411 415 415 441 393 394 424 371 399 446 407 433 412373 437 416 410 432 432 409 468 399 407 425 404 434 399 401402 401 446 393 388 445 459 369 409 429 441 390 439 436 438408 449 430 417 447 391 404 412 414 409 424 439 412 394 419428 408 426 401 442 428 412 425 408 425 434 423 400 444 388406 376 447 429 427 409 372 416 404 439 419 455 401 411 389399 431 435 410 393 435 407 439 398 433 454 443 431 426 397405 409 436 432 414 434 472 458 399 408 396 397 465 423 442446 421 447 423 452 440 420 412 404 459 438 432 415 422 402414 398 442 390 393 442 436 392 407 437 429 412 419 428 448426 386 397 381 449 418 437 384 429 389 418 434 455 452 380418 433 405 400 420 421 418 420 417 396 423 420 435 414 429
We use this data to test a hypothesis; nuclear decays are random processes, with afrequency that is a Poisson random variable. We could use a Normal PDF as well.To analyze this data, I ran it through binsort in the support software, which sorts the datainto 20 equivalence classes and predicts the correct Poissonian parameter (using the unbiased
15
estimator). Below we have the output.
Best-fit Poisson parameter is µ = 10.148333 Total count = 600, number of bins (classes) = 20
Bin j lower bound upper bound frequency mj theoretical freq. mj,th(mj−mj,th)2
mj,th
0 355.5 361.5 3 0.023485 377.2507731 361.5 367.5 3 0.238331 32.0009152 367.5 373.5 6 1.209332 18.9778333 373.5 379.5 6 4.090901 0.8909184 379.5 385.5 14 10.378958 1.2633205 385.5 391.5 22 21.065824 0.0414276 391.5 397.5 37 35.630501 0.0526387 397.5 403.5 42 51.655743 1.8048988 403.5 409.5 62 65.527462 0.1898909 409.5 415.5 49 73.888281 8.38328510 415.5 421.5 59 74.984291 3.40734811 421.5 427.5 64 69.178689 0.38767512 427.5 433.5 71 58.504033 2.66903313 433.5 439.5 64 45.670648 7.35626014 439.5 445.5 31 33.105783 0.13394415 445.5 451.5 27 22.397901 0.94559416 451.5 457.5 18 14.206336 1.01306117 457.5 463.5 12 8.480625 1.46050518 463.5 469.5 5 4.781345 0.00999919 469.5 475.5 2 2.553825 0.120103
To fit to a Normal curve, use the estimators for mean and standard deviation of the lastsection.Notice that when mtheor is small, such as for bins far from the most populated, the contri-bution to the χ2 statistic is huge. This is a well known phenomenon; the relative error in anessentially null measurement can be huge. Null measurements are to be avoided.
χ2 =∑19
j=0(mj−mj,th)2
mj,th= 458.359406
℘(χ2 ≥ 458.359406) =∫ ∞
458.359406
1
2182 Γ(18
2)x
182−1e−
x2 dx = 0.000000
16
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 210
7.1
14.2
21.3
28.4
35.5
42.6
49.7
56.8
63.9
71
Notice that the first three data bins really throw off the whole goodness of fit criterion, andjust for a few counts. If we discard this data, the result is the reduction of χ2 to χ2 = 30.1299,and we find then that
℘(χ2 ≥ 30.1299) =∫ ∞
30.1299
1
2182 Γ(18
2)x
182−1e−
x2 dx = 0.036486
If instead we sort into 15 equivalence classes, we see a better result
Best-fit Poisson parameter is µ = 7.488333 Total count = 600, number of bins (classes) =15
17
Bin j lower bound upper bound frequency mj theoretical freq. mj,th(mj−mj,th)2
mj,th
0 355.5 363.5 5 0.335745 64.7970431 363.5 371.5 5 2.514170 2.4578112 371.5 379.5 8 9.413470 0.2122383 379.5 387.5 19 23.497067 0.8606874 387.5 395.5 40 43.988468 0.3616375 395.5 403.5 56 65.880062 1.4817176 403.5 411.5 77 82.221978 0.3316527 411.5 419.5 76 87.957939 1.6256908 419.5 427.5 81 82.332296 0.0215599 427.5 435.5 97 68.503520 11.85412710 435.5 443.5 56 51.297719 0.43104211 443.5 451.5 40 34.921311 0.73860612 451.5 459.5 25 21.791868 0.47229113 459.5 467.5 8 12.552675 1.65119014 467.5 475.5 4 6.714187 1.097201
χ2 =∑14
j=0(mj−mj,th)2
mj,th= 88.394485
℘(χ2 ≥ 88.394485) =∫ ∞
88.394485
1
2132 Γ(13
2)x
132−1e−
x2 dx = 0.000000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 160
9.7
19.4
29.1
38.8
48.5
58.2
67.9
77.6
87.3
97
18
In this case discarding the first two bins reduces χ2 to χ2 = 21.13964, and using the chi-squared table generator in the support software, this gives
℘(χ2 ≥ 21.1396) =∫ ∞
21.1396
1
2132 Γ(13
2)x
132−1e−
x2 dx = 0.070014
We read this as saying that if the deviations from the best fit Poissonian of less than 7% areaccepted to be insignificance, we have a good fit, and our hypothesis is verified to be true;nuclear decays are random processes.
It should be clear that this experiment is sensitive to the size of the data set, the larger, thebetter. The analysis should be performed by regarding the contributions of dataclasses far from the most populated classes in the proper light. These classes willcontribute most to χ2 but will account for the smallest percentage of the actual data.
19
3 The Half-life of Ba137
The purpose here is to gain more experience with statistics and to learn how to use theGeiger counter.
3.1 Background
The Cesium sample will be used to determine the operating range (voltage) for the Geiger-Muller tube. The scaler-timer simply records decays produced by ionizing radiation emittedby the Barium and detected by the Geiger-Muller tube.
Radioactive Barium such as that used in this experiment decays by gamma ray emission,each decay releasing a gamma ray of 0.662 MeV
13756 Ba
m →13756 Ba+ γ
in which m denotes a meta-stable isotope of the element, and the right hand side containsthe stable isotope.Other types of nuclear decay modes are α decay
pmX →p−4
m−2 Y +42 He
in which an α particle or Helium nucleus is emitted, electron capture
p+ + e− → n0 + νe
in which a neutrino is produced with a neutron and β decay
p+ → n0 + e+ + νe
a decay mode of the proton in an unstable nucleus producing a positron and neutrino, andit’s inverse process
n0 → p+ + e− + νe
the common decay mode of the neutron.A typical decay chain for radioactive isotopes, in particular the one leading to the formationof Radon gas, is illustrated below.
20
We hypothesize that the probability of decay for a particle within time dt is
d℘ = λ dt
the rate at which the number of particles in a sample of size N is changing due to decay isthen
dN = −Nλdtwhich integrates to an exponential decay law
N(t) = N0e−λt
for the number of particles left after time t , beginning with N0. The time it takes for thesample to be cut in half by radioactive decays is
N0
2= N0e
λT 12
21
or
T 12
=ln 2
λ
We can compute the mean lifetime τ by finding the probability that a particle has notdecayed by time t. Since N(t) = N0e
−λt is the number remaining at time t, the fraction ofparticles that have not decayed is
f(t) =N(t)
N0
= e−λt
and so the mean lifetime is
τ =∫ ∞
0te−λtdt =
1
λ
which is what we will compute in the experiment. During the experiment we will not beable to directly measure the number of remaining nuclei, but rather the number of decays,N0 − N(t) which also drops with the same exponential rate. Let ∆N be the number ofdecays detected between t and t+ ∆t, then
∆N = N(t)−N(t + ∆t) = N0(1− e−λ∆t)e−λt = ∆N0e−λt
the quantity ∆N0 is fixed by the size of the interval ∆t and so we see that the actual numberof decays detected will also exponentially drop as the number of particles capable of decaydrops.
The standard measure of radioactivity of a sample is called it’s activity which is the numberof decays per second , one decay per second being called a Bq or Becqueral. We use theCi or curie instead
1Ci = 3.7× 1010Bq
The activity R(t) at a given time is
R(t) =−dNdt
= λN0e−λt = R0e
−λt
Since the activity is actually the decay rate, we see that our experiment will actually measurethis quantity, since we count for ten second periods, spaced by twenty second intervals. Thecount rate itself exponentially decays along with N .Our Barium sample has an activity of 9µCi. To put this into perspective, the Geiger-Marsden experiment was conducted with a 0.1Ci radium sample!
3.2 Apparatus
1. Geiger tube. Sargent-Welch, catalog number S-72095-85.2. Scaler-timer. Sargent-Welch catalog number S-72095-10.3. Cesium Sample. Amersham-Searle, 0.1 µc.4. Barium sample . Redco mini-generator, miniature radio-isotope generator, 9µc.
22
3.3 Procedure
The first step is to collect a Barium sample from the mini radio-isotope generator. An acidsolution is dripped into the generator which washes the surface of a Cesium/Barium sample,and the residue collects on a small disc of blotter paper placed below the generator. Bariumis produced from the Cesium by a beta decay
13755 Cs→136
56 Ba+ e− + νe
The Geiger counter operating voltage is now determined by counting the other radioactivesource, Cesium. The Cesium isotope used has a 27 year half-life and so provides a constantrate of decays for the counter.The Barium half-life is 2.6min.The Barium sample is placed under the counter and the experiment begins. We will countfor ten second intervals separated by twenty second periods. In other words, count for ten,rest for twenty, count for ten, rest for twenty and so on.
3.4 Data and analysis
Time Cnts per 10 sec0 13930 10360 8790 72120 60150 66180 41210 37240 41270 26300 22330 18360 19390 14420 11450 13480 11510 10540 6570 8600 5
We now need to correct for background radiation. Wait for one hour, and count thesample for 600 seconds. We find a total of 198 counts, or 3.3 counts per 10 second interval
23
can be attributed to background radiation. We now correct the data for this and work it upusing the awk script
#analyze.awk
print $1,"&",$2,"&",$2-3.3,"&",sqrt($2-3.3),"&",log($2-3.3),"\\\","\\hline"
This results in the table below
Time Cnts per 10 sec Cnts’ σCnts′ ln |Cnts′|0 139 135.7 11.649 4.9104530 103 99.7 9.98499 4.6021760 87 83.7 9.14877 4.4272490 72 68.7 8.28855 4.22975120 60 56.7 7.52994 4.03777150 66 62.7 7.91833 4.13836180 41 37.7 6.14003 3.62966210 37 33.7 5.80517 3.5175240 41 37.7 6.14003 3.62966270 26 22.7 4.76445 3.12236300 22 18.7 4.32435 2.92852330 18 14.7 3.83406 2.68785360 19 15.7 3.96232 2.75366390 14 10.7 3.27109 2.3702420 11 7.7 2.77489 2.04122450 13 9.7 3.11448 2.27213480 11 7.7 2.77489 2.04122510 10 6.7 2.58844 1.90211540 6 2.7 1.64317 0.993252570 8 4.7 2.16795 1.54756600 5 1.7 1.30384 0.530628
withCnts′ = Cnts− Cntsback
is the corrected number of counts per 10 second interval,
σCnts′ =√Cnts′
is the standard deviation in the number of counts. We can compute the error in the log ofthe number of counts
∆ lnCnts′ =∆Cnts′
Cnts′=σCnts′
Cnts′=
1√Cnts′
and produce a table of file of data suitable for error-bar graphing with the awk script
24
#analyse2.awk
print $1," ",log($2-3.3), " ", 1.0/sqrt($2-3.3)
Time ln |Cnts′| σln |Cnts′|
0 4.91045 0.08584430 4.60217 0.1001560 4.42724 0.10930490 4.22975 0.120648120 4.03777 0.132803150 4.13836 0.126289180 3.62966 0.162866210 3.5175 0.17226240 3.62966 0.162866270 3.12236 0.209888300 2.92852 0.231249330 2.68785 0.26082360 2.75366 0.252377390 2.3702 0.30571420 2.04122 0.360375450 2.27213 0.321081480 2.04122 0.360375510 1.90211 0.386334540 0.993252 0.608581570 1.54756 0.461266600 0.530628 0.766965
The graph of which is below, created by lg in the support software. You could use gnu-plot which has error bar graphing.
25
0 51 102 153 204 255 306 357 408 459 5101
1.4
1.8
2.2
2.6
3
3.4
3.8
4.2
4.6
5
With log of corrected counts of the vertical axis and time in seconds on the horizontalaxis. We can run a linear regression on the data , but first we will discard the three datapoints indicated in the table with an asterisk. These arepoints with count totals so close tothe background level that they are most probably overwhelmed by the fluctuations in thebackground count rate. It is questionable scientific procedure to discard data, unlessthere is very good reason. The linear regression program in the appendix gives us a slopefor these 17 points of
y = 4.811657 +−0.005963x, χ2 = 0.114469, σ2m = 0.000002
which results in
m = −0.005963± 0.001414, τ 12
= − ln 2
−0.005963 1s
= 116 s
which places the half-life between the limits
ln 2
|m+ σm|≈ ln 2
|m| (1 +σm
|m| + · · ·) = 152.28 sec
andln 2
|m− σm|≈ ln 2
|m| (1−σm
|m| + · · ·) = 93.95 sec
26
which is in fair agreement with the accepted 2.6 min = 156 sec value. When we perform thisexperiment we should be sure that enough isotope is collected that we have at least 600or more counts initially. In this way we will have enough counts at subsequent times thatbackground fluctuations do not overwhelm our counts.
3.5 Background material; the Nuclear Shell Model
The actual potential felt by nucleons in a nucleus is complex and the Schroedinger equationis impossible to solve analytically. However for deeply bound nucleons the potential looks alot like a three dimensional oscillator and the nuclear shell model is based on this potential.It is very useful for predicting stable nuclei configurations.Begin with the nonrelativistic, three dimensional oscillator in polar coordinates. Since thepotential is spherically symmetric we find
[H, ~L2] = [H,Lz] = 0
and so we decompose the wavefunction into a radial part and a spherical harmonic
ψ = R(r)Y`,m(θ, φ)
and the radial equation becomes
(d2
dr2=
2
r
d
dr+
2m
h2 (E − h2`(`+ 1)
2mr2− mω2r2
2))R(r) = 0
Perform the variable change
ξ =mωr2
h2
d
dr= 2
√
mω
h2
√
ξd
dξ
and we find
(ξd2
dξ2+
3
2
d
dr+ (
ξ
2hω− `(`+ 1)
4ξ− ξ
4))R(ξ) = 0
We now impose the boundary conditions on the wavefunction
R(0) = 0
R(∞) = 0
by redefining the wavefunction
R(ξ) = ξl2 e−
ξ2u(ξ)
and we find that
27
(ξd2
dξ2+ (
3
2− `− ξ) d
dξ+ (
E
2hω− 3
4− `
2))u = 0
and again we recognize the same confluent hypergeometric equation
x f ′′(a; c, x) + (c− x)f ′(a; c, x)− af(a; c, x) = 0
that has appeared in every potential problem that we have studied in Physics 441. We knowthat a series solution exists
f(a; c, x) =∞∑
j=0
(a)j
(c)j
xj
j!, (a)j = a(a+ 1) · · · (a+ j − 1)
which will truncate at a polynomial (satisfying the condition at ∞) if
−a =E
2hω− 3
4− `
2= N
where N = 0, 1, 2, 3, · · ·. The spectrum that we obtain is
E = hω(2N + `+3
2)
Each ` value has 2`+1 degenerate m` values and a little arithmetic shows that the degeneracyof a level of energy
E = hω(M +3
2)
with M = 0, 1, 2, · · · is
D(M) =(M + 1)(M + 2)
2
For a fixed value of M , ` can be M , M − 2, M − 4,· · ·, however ` must remain positive. Theplot of energy versus angular momentum forms straight lines called Regge trajectories.
You will often see this written in the literature in terms of the oscillator radial quantumnumber n as
n =E
2hω− `
2+
1
4
with solutions that can be written in terms of the Laguerre polynomial, which is a hy-pergeometric function
Ψn,`,m(r, θ, φ) = An,`e− 1
2ρ2
L`+ 1
2n−1(ρ
2)Y`,m(θ, φ)
with
ρ =r
r0, r0 =
√
h
mω, ξ = ρ2
28
and the Laguerre polynomial satisfies the recursion
Lmk (x) =
1
k!ex x−m dk
dxk(xk+m e−x)
These functions are orthonormal on [0,∞] with respect to kernel xm e−x;
∫ ∞
0xm e−x Lm
k (x)Lmk′(x) dx =
Γ(m + k + 1)
k!δk,k′
which will allow one to show that the normalization factor for the wavefunctions is
An,` =
√
√
√
√
2n+`+1 (n− 1)!
(2n+ 2`− 1)!!√π r3
0
We now try to find the oscillator frequency ω that best fits the experimental data as far asthe size of the nucleus and the binding energy of a nucleon in the nucleus is concerned. Todo this we compute
∫
Ψ∗n,`,m r
2 Ψn,`,m d3x = 〈r2〉
and fit this to the experimental radius-squared of a nucleus containing a total of A nucleons.This computation is a good excerise in quantum mechanics and the various recursion relationssatisfied by the Laguerre polynomials, and results in
〈r2〉 = (2n+ `− 1
2) r2
0
and from the rather extensive experimental data, we have the result that the nucleus con-taining A nucleons subjects its nucleons to a potential pretty well described by a harmonicoscillator with
hω =34.7MeV
A13 − 0.44
, rr = 1.19√
A13 − 0.44 fermi
This is pretty useful for computing γ ray energies for nuclei that decay through γ emission.
Analogous to noble gasses with filled atomic shells, in the nuclear shell model we assumethat nuclei are stablest when they have nucleons filling each shell . Each state
|N, `,m` >
can contain two protons and two neutrons, one spin up,one spin down of each. This givesrise to stable nuclei of atomic number (Z)
2 = Z
2 + 6 = 8 = Z
2 + 6 + 2 + 10 = 20 = Z
29
2 + 6 + 2 + 10 + 6 + 14 = 40 = Z
These are called Magic Number nuclei. Indeed the Z = 2, 8, 20 nuclei are known to bevery stable. After that the nucleon spin-orbit coupling disrupts the energy levels and breaksthe degeneracy, and alters the magic number pattern. We define the term symbol of a singlenucleon in a way very similar to that of an electron in an atom, with nomenclature
(N + 1)Lj = nLj
and the nuclei of various elements are constructed by a nuclear Aufbau process; we beginadding nucleons to a shell model manifold of single particle states, until we reach Z protonsand A− Z neutrons.Even better predictions of the magic numbers A of stable nuclei come from including theeffects of the spin-orbit interaction beteen nucleons, altering the potential to
V (r) = −V0 +1
2mω2r2 − 2α
h2~L · ~s
The eigenstates of the full Hamiltonian become
Ψn, `, j,mj(r, θ, φ) = An,`e− 1
2ρ2
L`+ 1
2n−1(ρ
2)Vj=`± 12,`,mj
(θ, φ)
in which the angular part of the wavefunction is an eigenstate maximal set of commutingoperators derivable from the total angular momentum operator
~J = ~L+ ~s = ~L +1
2h~σ
containing contributions from nucleon orbital and spin angular momenta.You hopefully remember that that means
JzVj,`,mj= hmj Vj,`,mj
and( ~J)2Vj,`,mj
= h2j(j + 1)Vj,`,mj
from quantum theory. But( ~J)2 = (~L)2 + 2~L · ~s+ ~s2
and so
~L · ~s =j(j + 1)− `(`+ 1)− s(s+ 1)
2
and by the rules of angular momentum addition, since s(s+ 1) = 12(1
2+ 1),
j = `+1
2, or j = `− 1
2
We find then that the energy En,j,` of these states will be
En,`+ 12,` = (2n+ `− 1
2)hω − V0 − α`
30
and
En,`− 12,` = (2n+ `− 1
2)hω − V0 + α(`+ 1)
This results in the single-nucleon states shown below, which is in very good agreement withexperimentally established stable nuclei-nuclei with filled shell-model shells.The state of lowest energy in the shell model is
nLj = 1s 12, E1, 1
2,0 =
1
2hω − V0
and the table below together with the formula for ω can be used to determine nuclear energylevel spacings, and the frequencies of γ rays emitted when a nucleon changes levels.
n ` State En,j,` − E1, 12,0 Nj
1 0 1s 12
0 2
first magic filled shell; 2 particles1 1 1p 3
2hω − α 4
1 1 1p 12
hω + 2α 2
second magic filled shell; 8 particles1 2 1d 5
22hω − 2α 6
2 0 2s 12
2hω 2
1 2 1d 32
2hω + 3α 4
third magic filled shell; 20 particles1 3 1f 7
23hω − 3α 8
fourth magic filled shell; 28 particles2 1 2p 3
23hω − α 4
2 1 2p 12
3hω + 2α 2
1 3 1f 52
3hω + 4α 6
1 4 1g 92
4hω − 4α 10
fifth magic filled shell; 50 particles
The superstable magic filled shells are well separated in energy from the next highest set of(empty) one particle states.Furthermore, experimental data shows that α depends on `, but is small, for example (inMeV);
αn,`;α1,2 = 1.15, α1,3 = 0.93, α1,4 = 0.76
α2,1 = 0.68, α2,2 = 0.58, α3,1 = 0.19
and it continues to decrease for both increasing n and `.An energy level diagram similar to that used to perform the Aufbau process for atomic statesis illustrated below.
31
Remember that we fill in neutrons and protons independently.
Example Apply the nuclear Aufbau process to build up the most stable Helium isotope.Such an isotope will have a filled shell for both types of nucleons.
32
1s
1p
1p
1d
1/2
1/2
3/2
5/2
-1/2 1/2 -1/2 1/2
protons neutrons
In order to use the shell model, which describes only single nucleon states, to describe real,multinucleon nuclei, we use the Aufbau process to add nucleons to the shell model statescomputed with the appropriate A, and further take into acount the fact that paired, likenucleons have an additional pairing energy of
Epairing = 211.2MeV√
A
By pairing we mean; the two combine angular momenta to 0. Remember that two parti-cles of angular momentum j can have two-particle wavefunctions of angular momentum 2j,2j − 1, 2j − 2, · · · , |j − j| = 0. This pairing increases the binding energy of the resultantstate.For a filled shell, this pairing amounts to the combination of all angular momenta of theshell’s nucleons to j = 0. A filled shell has no contribution to the nuclear spin.
This means that we need to distinguish between three nuclear types;even-p, even-n; or Z even, N = A−Z even, so all protons are paired and all neutrons arepaired.A-even, Z-odd; or Z odd, N = A − Z odd, so all protons but one are paired and allneutrons but one are paired. This is also called odd-odd.odd-even nuclei; Z even, N odd or Z odd, N even.The pairing is with respect to mj, so we expect pairing between |n, j, `,mj〉 and |n, j, `,−mj〉.
The shell model can be used to determine properties of the entire multinucleon nucleus,which are going to be largely determined for odd-even nuclei by the last unpaired nu-cleon.
Predictions;The spin of an even-even nucleus in the ground state should be zero. This conforms toexperimental evidence.Due to cancellations by the pairing, the nuclear spin of odd-even nuclei is determined by
33
the wavefunction of the last nucleon, the ground state nuclear spin being the angularmomentum j of the last unpaired nucleon.Several nucleons in an incomplete shell often do combine their angular momenta to producea different value J 6= j. Suppose that there are k ≤ 2j + 1 nucleons in a shell, they mayproduce a state whose label is
(n(`)j)kJ
and a perfect example is the Sodium isotope used in several of our experiments
Ne2110, Na2211 are in state (1d 5
2)3
32
Some odd-even nucleiZ A− Z A Name J Last unpaired nucleon1 0 1 H 1
21s 1
2
3 4 7 Li 32
1p 32
5 6 11 B 32
1p 32
9 10 19 F 12
2s 12
11 12 23 Na 32
1d 52
19 20 39 K 32
1d 32
19 22 41 K 32
1p 32
25 30 55 Mn 52
1f 72
27 30 57 Co 72
1f 72
27 32 59 Co 72
1f 72
Example O168 is doubly magic, and superstable.
1s
1p
1p
1d
1/2
1/2
3/2
5/2
-1/2 1/2 -1/2 1/2
protons neutrons
-3/2 -1/2 1/2 3/2
-1/2 1/2
-3/2 -1/2 1/2 3/2
-1/2 1/2
O157 is doubly magic except for a proton hole in 1P 1
2. Since in a filled shell all nuclear spins
are paired, this hole has spin one half, and therefore so does this nucleus. We define the jvalue of a filled shell to be zero, a consequence of the nucleon urge to pair. This means that anucleus will have the quantum numbers of holes in filled shells. The N 15
7 does experimentally
34
have j = 12. The nucleus O17
8 is doubly magic except for an extra nucleon in the 1D 52
and so
the shell model predicts a nuclear spin j = 52
again in agreement with experiment. Nuclearparity under inversion of the coordinate system is given by
P = (−1)(∑
(nucleon ` values)
for example the nucleus
O168
has a filled s shell , each nucleon having ` = 0, and a filled p shell with each nucleon having` = 1 and so has even parity, a fact verified experimentally. Nuclear spin and parity actuallyplays a very imporant role in the spectra of diatomic molecules, and the parity of the wave-function will determine which rotational modes exist for the molecule. This will be evidenteven in the thermodynamic functions for gases of such molecules.
We can use the concept of pairing energy to understand β decay processes of some of thenuclei used in this course, such as Na22
11. Consider a Na2211 with an unpaired proton and
neutron. Remember that the neutrons are bound a little tighter than the protons due thethe raising of the proton energy caused by the Coulombic repulsion of like charged particles.The total electrostatic potential energy due to a nuclear charge Z is
Ucoul =1
24π∫ r0
0r2(
Zr
4πε0r30
)2 dr +1
24π∫ ∞
r0
r2(Z
4πε0r2)2 dr =
3
5
Z2
4πε0r0
in which r0 is the mean nuclear radius. Using the shell model value we see that this is
Ucoul =3
5
Z2
4πε0 (1.21F )A13
=0.714Z2MeV
A13
which evaluates to 30.83MeV for Na2211.
The most likely decay therefore, in order to achieve pairing, is for the unpaired proton toinverse-β decay
1s
1p
1p
1d
1/2
1/2
3/2
5/2
-1/2 1/2 -1/2 1/2
protons neutrons
-3/2 -1/2 1/2 3/2
-1/2 1/2
-3/2 -1/2 1/2 3/2
-1/2 1/2
-5/2 -3/2 -1/2
β+
ν
1s
1p
1p
1d
1/2
1/2
3/2
5/2
-1/2 1/2 -1/2 1/2
protons neutrons
-3/2 -1/2 1/2 3/2
-1/2 1/2
-3/2 -1/2 1/2 3/2
-1/2 1/2
-5/2 -3/2 -5/2 -3/2 -1/2 1/2
γ
35
A proton first decays to a neutron, emitting a positron that quickly captures an electron fromthe innermost shell (K shell) of the electron cloud, the resulting annihilation produces twoγ’s at 0.511MeV that are not shown in the figure. Recall that the total nuclear spin of theSodium Na22
11 was 3 before the decay. Remember that filled shells have nucleon spins pairedto zero, and in the valence shell there is one unpaired proton and one unpaired neutron withjp = jn = 5
2, and so the nuclear spin is one of the spins in the decomposition
5
2⊗ 5
2= 5⊕ 4⊕ 3⊕ 2⊕ 1⊕ 0
Conservation of angular momentum requires that instantly after the decay, it be 3±1, in factit is experimentally found to be 2. The decay leaves a Neon nucleus with an even numberof protons and neutrons, which now quickly pair, leaving a spin-0 nucleus and shedding theexcess energy as another γ. This line of reasoning is probably qualitatively correct, but Ihave not been able to compute the energy of the last γ-emission correctly from the shellmodel.A useful source of information is Nuclear Forces by Gernot Eder, MIT Press, 1968.
4 Determination of the Speed of Light
At one time the speed of light was thought to be infinite. A correct determination of thespeed of light is a crucial experiment in modern physics and can be performed with relativelysimple apparatus.
Laser120 V AC
Transformer120 V AC
Photoelectric tube
Frequencymeter
120 V AC
Meter stickscale
Beam splitter
Fixed mirror
Rotatingmirror
Powersupply
36
Apparatus
1. Power supply. Spectra Physics laser exciter model #249.2. Laser. Spectra Physics Stabilite Helium-Neon laser model #120.This is a powerful laser. Do Not Look Into The Beam!3. Beam splitter. A glass slide.4. Rotating mirror. Leybold 476-41.5. Photoelectric tube.6. Transformer. Variac type W5M73.7. Flat mirror and stand.8. Magnifier lens.9. Frequency meter. Fluke 1900A multi-counter.
The rotating mirror is shown below.
The apparatus is set up with about 5 meters between focusing lens and laser. The distancefrom the meter stick to the beam splitter is the same as from beam splitter to rotatingmirror. In this case the emitted beam from the laser and reflected beam from the splitter tothe meter stick scale have the same path length, and the returning beam reflected from thesplitter will focus to a point more or less on the scale. The Photoelectric tube must be set upin such a position that each time the rotating mirror undergoes a single rotation, it reflectsthe laser beam into the photo-tube at some point, and the number of pulses per second isthe inverse of the rotation frequency of the rotating mirror. This may require that the beamsprayed into the tube be out of line with the beam reflected back into the beam splitter toavoid interference of the two beams.
Consider the figure below, showing two beams arriving at the scale, one bounces off ofthe rotating mirror and heads directly to the scale. The current position of the mirror isposition 2. At an earlier time, the mirror was at position 1, and the beam (very finely dottedbeam) from the laser was directed by the rotating mirror towards the distantsecond mirror.Now it is returning and hits the mirror in it’s current position but does so at an angle suchthat it is not directed back along it’s original path. These two beams arrive at the scaleat the same time, but were emitted by the laser at different times separated by the time ittakes the light to travel back and forth between rotating and fixed mirrors. by an angle.
37
2θ
L
S
position 1
position 2
θ
S=distance between beam traces on scaleL=distance between fixed and rotating mirrorsD=distance from rotating mirror to laser.d=distance from laser to beam splitter=distance from beam splitter to scaleθ= change in position of mirrorν=rotation frequency of mirror in Hzω= natural rotation frequency of mirror.We have the basic relations
ω = 2πν, θ = ωdt, dt =2L
c
and
tan 2θ =S
D
so that
c =4π LD
Sν
Data
In this run we measured d five times and obtained
d = 53.1± 0.03cm
and measured D five times finding
D = 309.9± 0.5cm
38
and finally L five times obtaining
L = 3111.0± 3.0cm
When the separation S = 0.2 cm it was found that the rotating mirror had frequency 502 Hzand so
c =4π(3111 cm)(309.9 cm)(502 Hz)
0.2 cm= 3.041× 108m
s
which is not far off of the accepted value of c = 2.9979× 108 ms.
Things to watch out for. Calculate ν correctly; the mirror is two-sided!
39
5 Determination ofCp
Cvfor a gas
The specific heat ratio for a gas is known as the adiabatic constant and can be measured bystudying a typical adiabatic process such as sound wave propagation in the laboratory.
5.1 Background
Sound is a longitudinal compression wave in a material medium such as a gas or solid. Wewill first derive the wave equation in order to find the velocity of sound and the relationbetween the displacement and pressure waves.Consider a layer of molecules of gas of cross sectional area A and thickness dx.
x x+dx x+ x+dx+d η η+ η
AA
P(x)
P(x+dx)
Let passage of the wave through the gas cause the left wall to displace by an amount η tothe right and the right wall of the layer displace by η+dη. This will cause the cell to expandand so the air density and pressure inside of the cell will change. The old cell volume was
V0 = Adx
and the mass of the gas in the cell was and still is
dm = ρ0Adx
in which ρ0 is the density of the gas at equilibrium. The new cell volume is
V0 + ∆V = A(dx+ dη)
and so the volume change is
∆V = Adη
and we find that
40
∆V
V0=dη
dx
The force causing these displacements is supplied by pressure gradients, the pressure onthe left wall of the cell, P (x), pushes to the right and that on the right wall
P (x+ dx) = P (x) +∂P
∂xdx+ · · ·
pushes to the left. We find that the net force to the right is
Fx = A(P (x)− P (x+ dx)) = −Adx∂P∂x
Since the cell has acceleration
ax =d2η
dx2
the equation Fx = max reads
ρ0dxAd2η
dx2= −Adx∂P
∂xor
d2η
dx2= − 1
ρ0
∂P
∂x
This is the acoustic equation. We need to establish the thermodynamics of the process,after all we are dealing with the expansion of an ideal gas presumably. Sound propagationis a fast process, and no heat is exchanged between adjacent gas cells and so propagation ofsound is adiabatic. In that case
P0Vγ0 = (P0 + ∆P ) (V0 + ∆V )γ
in which γ is the ratio of specific heats
γ =Cp
Cv
We can expand the right hand side using the Binomial theorem
(1 + δx)n = 1 + nδx+ · · ·for small x, which in our case becomes
(V0 + δV )γ = V γ0 (1 +
δV
V0
)γ
= V γ0 (1 + γ
∆V
V0+ · · ·)
and if we again neglect the product ∆V δP we find that
41
∆P = −γ∆V
V0P0 = −γ P0
∂η
∂x
and the total gas pressure is
P (x) = P0 + ∆P = P0 − γP0∂η
∂x
Insertion of this into the acoustic equation gives all the same results as before except nowwe find that sound waves travel (adiabatically) at speed
v =
√
γP0
ρ0=
√
γRT
M
where M is the molecular weight in grams per mole.A full treatment of the specific heats CP and CV for polyatomic gasses is given at the endof this experiment.
5.2 Apparatus
1. Kundts tube apparatus.2. High pressure gas cylinders of Ar, N2, CO2.3. Oscilloscope.4. Audio oscillator and speaker.5. Thermometer.
This apparatus is essentially ”canned” and lives in the physical chemistry laboratory.
AudioOscillator
Oscilloscope
Speaker
Thermometer
Gas OutGas In
Moveable PistonRod
Microphone
Kundt’s tube
Meter stick or scale
5.3 Procedure
An oscilloscope will be used to monitor the phase relationship between the input sound signalinto the gas in the tube and the signal picked up by a microphone. Such a phase relation
42
can be displayed as a Lissojous figure by displaying one signal on the horizontal input, andthe other on the vertical input to the scope. Typical Lissojous figures for example with
y(t) = Ax sin(ωt), y′(t) = Ay sin(ωyt− φ)
look like the following
This will require that the apparatus be calibrated by holding a tuning fork near the mi-crophone with the speaker disconnected from the scope input. The apparatus needs to becalibrated at several frequencies around 1 to 2 kilohertz. The program used to create thesefigures is included at the end of this experiment.
The piston should be positioned near the end of the tube as seen in the figure, and thetube flushed for about 10 minutes with the gas to be studied. Before data is ready to becollected, reduce the flow rate to a low value, but not zero, to prevent air from diffusing intothe tube, which is not sealed. Adjust the gain on the scope until both input and microphonesignals are at the same magnitude.
Set the audio oscillator on the desired frequency, 1 kHz for Nitrogen and Carbon dioxide,2 kHz for He or Ar, and slowly move the piston towards the speaker end of the tube. Whena half wave is standing in the tube the waveform being input will be
y(x, t) = A sin2πx
λ
while the microphone receives
y′(x, t) = A′ sin2π(x+ λ
2)
λ= −A′ sin
2πx
λ
eliminating x we should expect the curve
y =A
A′ y′
or a straight line to be displayed on the scope. These runs need to be repeated and statisticalanalysis performed.
43
5.4 Data and analysis
This data was taken by a student in spring 1997.
λ2
in cmTrial Air CO2 Ar
1 17.40 13.65 16.102 17.25 13.45 16.103 17.20 13.55 16.204 17.50 13.60 16.055 17.10 13.55 16.106 17.35 13.45 16.05
Ave. 17.30± 0.15 13.54± 0.08 16.10± 0.05
All measurements were taken at temperature
T = 25.0C = 298K
and at frequencyν = 1000.0Hz
Using the ideal gas constant
R = 8.31J
moleoK
we obtain
Quantity Air CO2 Arλ2, cm 17.30± 0.15 13.54± 0.08 16.10± 0.05v, m
s346.0± 3.0 270.8± 1.6 322.0± 1.0
M, gm
mole28.97 44.01 39.948
γ = Cp
Cv1.400± 0.024 1.303± 0.015 1.673± 0.01
44
The exact values are as follows. Diatomic molecules have one vibrational, two rotationaland three translational degrees of freedom., monatomics have only three translational, andCarbon dioxide has vibrational modes, you should read the next section thoroughlyin order to see where the theoretical values come from.
Quantity Air CO2 Ar
γexp 1.400± 0.024 1.303± 0.015 1.673± 0.01γtheory 1.400 1.285 1.667
% difference 0.0% 1.4% 0.38%
5.5 Molecular spectra and the equipartition theorem
The precise form of both CP = (∂U∂T
)P and CV = (∂U∂T
)V in a quantum mechanically correctdetermination are both temperature dependent, both look like “steps” or plateaus, indicat-ing that certain molecular motion modes are excited at various temperature regimes. Whatmodes are we talking about? There are three types of molecular motions; translations androtations (purely kinetic), and vibrations (possessing both kinetic and potential energy).We can compute the specific heat of a molecule with classical formalism for a simple illus-trative case such as the linear molecule CO2. Consider then the lagrangean for the moleculefixing all atoms to lie on a line;
L =1
2mC z
2 +1
2mO(z − x1)
2 +1
2mC(z + x2)
2 − 1
2k(x1 − `)2 − 1
2k(x2 − `)2
in which z fixes the position of the Carbon in space, and xi is the distance between eachOxygen and the Carbon. The equations of motion are
d
dt(∂L∂z
) =∂L∂z
ormC z + 2mOz −mOx1 +mOx2 = 0
d
dt(∂L∂x1
) =∂L∂x1
ormOx1 −mOz = −kx1
and finallyd
dt(∂L∂x2
) =∂L∂x2
ormOx1 +mOz = −kx2
Following the usual T and V matrix methods for normal modes, we can write this as
mC + 2mO −mO mO
−mO mO 0mO 0 mO
zx1
x2
=
0 0 00 −k 00 0 −k
zx1
x2
45
We assume that each coordinate is an amplitude times eiωt, which results in a secular equationfor the amplitudes and frequencies
−(mC + 2mO)ω2 mOω2 −mOω
2
mOω2 −mOω
2 + k 0−mOω
2 0 −mOω2 + k
Az
Ax1
Ax2
= 0
The frequencies are determined by the vanishing of the determinant
0 = −(mC + 2mO)ω2(−mOω2 + k)2 − (−mOω
2 + k)(mOω2)2
This gives us three frequencies;
ω20 = 0, ω2
1 =k
mO
, ω22 =
(mC + 2mO)
mC +mO
k
mO
In terms of these normal modes, we could perform a coordinate transformation
ξ0ξ1ξ2
=
O1,1 O1,2 O1,3
O2,1 O2,2 O2,3
O3,1 O3,2 O3,3
zx1
x2
in terms of which the classical lagrangian would be
L′ =1
2ξ20 +
1
2ξ21 −
1
2ω2
1ξ21 +
1
2ξ22 −
1
2ω2
2ξ22
The classical Hamiltonian will be
H′ =1
2p2
0 +1
2p2
1 +1
2ω2
1ξ21 +
1
2p2
2 +1
2ω2
2ξ22
which results in the classical partition function
Z =∫
dp0 dp1 dp2 dξ0 dξ1 dξ2h3
e−βH′
, β =1
kT
It is a good exercise to compute this function by doing the integrals, each is from −∞ to∞, to compute the average energy of such a molecule in contact with a heat reservoir attemperature T
U = −∂ lnZ∂β
=5
2kT
and the specific heat at constant volume
CV = (∂U
∂T)V =
5
2k
In a one dimensional world, we find that our gas of such molecules will have
γ =CP
CV
=CV + k
CV
= 1.4
46
The Equipartition theorem is essentially just a rule of thumb gotten by inspecting theHamiltonian written in normal mode coordinates; CV gets a contribution of 1
2k from
each quadratic term in H′. In this case there are five terms.
In three dimensions, we will need three coordinates (x, y, z) to specify the position of theCarbon in space, the two bond distances, and two Euler angles θ and φ to specify theorientation of the molecule in space. In that case the classical Hamiltonian will be (innormal mode coordinates)
H′ =1
2~P 2 +
1
2I(P 2
θ +P 2
φ
sin2 θ) +
1
2p2
1 +1
2ω2
1ξ21 +
1
2p2
2 +1
2ω2
2ξ22
Counting quadratic degrees of freedom, we find nine, and so CV = 92k, and CP = CV + k
making the adiabatic constant 1.2222.
In the real world we find that stiff molecular vibrational modes are generally not excitedunless the temperature is quite high. This can only be seen from a quantum mechanicaltreatment of the problem. If we toss out the four vibrational degrees of freedom, we are leftwith five excitable molecular modes at low (room) temperature. This gives us γ = CP
CV= 7
5=
1.4 for CO2. The full quantum treatment of only the vibrational modes of this moleculewill give us an average energy
U =5
2kT +
hω1
2 sinh hω1
2kT
+hω2
2 sinh hω2
2kT
since the quantum Hamiltonian will be
H′ =1
2~P 2 +
1
2I(P 2
θ +P 2
φ
sin2 θ) + hω1(n1 +
1
2) + hω2(n2 +
1
2)
and the partition function for the quantized oscillation terms will be
∞∑
ni=0
e−βhωi(ni+12)
(sums rather than integrals). The rotational modes also will have a temperature dependence,an issue that we will not go into here. The corresponding specific heat will be
CV =5
2k +
h2ω21
4kT 2
cosh hω1
2kT
sinh2 hω1
2kT
+h2ω2
2
4kT 2
cosh hω2
2kT
sinh2 hω2
2kT
This has the adversised plateau structure, here plotted for ω2 = 6.0 · ω1;
47
0 2 4 6 8 100
1
2
3
4
5
2πkT/h
CV/k
You can see that the specific heat is roughly constant at some multiple of k, as classicallypredicted, but only in certain temperature regimes. This is a clear indication that at suffi-ciently low temperatures, the oscillatory modes are not excited and cannot absorb heat.
It turns out that CO2 has an additional oscillatory mode, a flexing out of the linear ge-ometry, that has a much lower frequency (because of weal restoring forces) than eitherbond-stretching mode. This comes out of a group theory treatment of molecular vibrations.If we work at a low enough temperature that all three translational modes, both rotational,and the flexing mode (Potential and kinetic degrees) are excited, the classical prediction forCO2 will be CV = 7
2k, and the adiabatic constant is then γ = 9
7= 1.2857, which is reported
as the theoretical value in the writeup. You should verify that for Ar, a monatomic, CV = 32k
and γ = 53
= 1.666, and for N2 or O2, if we neglect the vibrational modes of the extremelyrigid double bonds, CV = 5
2k and γ = 7
5= 1.4.
48
5.6 Safe Handling of High Pressure Gas Cylinders
GasCylinder
Shutoffvalve
High pressuregauge
Low pressuregauge
Regulator
Outlet valve
Gas hose
Regulatorhandle
Locate the shutoff valve on top of the cylinder, the regulator handle and outlet valve. Be surethe outlet valve is closed firmly but never forced. Set the regulator to zero flow by turningthe regulator handle counter-clockwise until you feel resistance diminish. If it is already setto zero flow it will turn freely. Open the shutoff valve. The high pressure gauge will nowread tank pressure.
Slowly turn the regular handle clockwise, screwing it in, until the low pressure gauge risesslightly above zero. This amount will be adequate for the experiment.Now turn the outlet valve. Even if the low pressure gauge reading drops there is still gas flow.
When you are done, first turn off the outlet valve (close it), set the regulator to zero, thenclose the shutoff valve. Open the outlet valve,then the regulator slightly, allowing gas trappedin the regulator to escape. The high pressure gauge should drop to zero. Once again closethe regulator and outlet valve.
Gas cylinders should always be secured to the workbench, and never use tools to force openor close a valve.
5.7 Lissojous program
/* Draw lissojou figures for illustration */
/* gcc lissojou.c -L/usr/X11R6/lib -lplot -lXaw -lXmu -lXt -lSM -lICE -lXext -lX11 -lm */
49
#include<stdio.h>
#include<strings.h>
#include<stdlib.h>
#include<math.h>
#include<plot.h>
#define N 1000
#define PI 3.1415927
float X[N],Y[N]; /* f[n] is a sine function for cuts */
float dt,t,Ax,Ay,Wx,Wy,phi;
int n,m,k,l;
int pl_handle;
char buffer[6],*labels,cmd[3],*cmdptr;
main (int argc, char *argv[])
char Axl[14]="A\\sbx\\eb=";
char Ayl[14]="A\\sby\\eb=";
char Wxl[14]="\\*w\\sbx\\eb=";
char Wyl[14]="\\*w\\sby\\eb=";
char phil[6]="\\*f=";
dt = 2.0*PI/(1000.0);
if(argc !=7)
printf("./lissojou Amp_x omega_x Amp_y omega_y phase display\n");
exit(0);
Ax=atof(argv[1]);
Wx=atof(argv[2]);
Ay=atof(argv[3]);
Wy=atof(argv[4]);
phi=atof(argv[5]);
cmdptr=argv[6];
pl_handle = pl_newpl (cmdptr, stdin, stdout, stderr);
pl_selectpl (pl_handle);
if (pl_openpl () < 0) /* open Plotter */
fprintf (stderr, "Couldn’t open Plotter\n");
return 1;
pl_fspace (-4.0, -4.0, 4.0, 4.0);
pl_flinewidth (0.005);
pl_erase();
pl_filltype(0);
pl_pencolorname ("black");
50
pl_fline(0.0, -3.8,0.0,3.8);
pl_fline(-3.8,0.0,3.8,0.0);
for(n=0;n<1000;n++)
t=(float)n*dt;
X[n]=Ax*sin(Wx*t);
Y[n]=Ay*sin(Wy*t-phi);
for(n=0;n<999;n++)
pl_fline(X[n+1], Y[n+1],X[n],Y[n]);
pl_fline(X[999], Y[999],X[0],Y[0]);
/* labels */
pl_ffontname("HersheySerif");
pl_ffontsize(0.25);
pl_fmove(3.4,-0.25);
pl_label("y(t)");
pl_fmove(0.15,3.7);
pl_label("y’(t)");
pl_fmove(-3.5, -3.5);
labels=gcvt(Ax,4,buffer);
strcat(Axl,labels);
pl_label(Axl);
pl_fmove(-3.5, -3.8);
labels=gcvt(Ay,4,buffer);
strcat(Ayl,labels);
pl_label(Ayl);
pl_fmove(-2.0, -3.5);
labels=gcvt(Wx,4,buffer);
strcat(Wxl,labels);
pl_label(Wxl);
pl_fmove(-2.0, -3.8);
labels=gcvt(Wy,4,buffer);
strcat(Wyl,labels);
pl_label(Wyl);
pl_fmove(0.15, -3.5);
labels=gcvt(phi,4,buffer);
strcat(phil,labels);
pl_label(phil);
pl_closepl ();
pl_selectpl (0);
51
pl_deletepl (pl_handle);
52
6 Electron Diffraction and Bragg Scattering
The purpose is to not only measure typical inter-atomic spacings between atoms in a typicalcrystalline solid (graphite) but to study as well the wave properties of electrons. The crystallattice will form a diffraction grating from which electronic waves will scatter.
6.1 Apparatus
1. Power supply A; a Teltron limited kV power supply, model 813. This consists of both a0− 5000 volt DC and a 6.3 volt AC supply.2. Power supply B; a Hewlett-Packard 0− 60 volt DC power supply, model 6218A.3. A Keithly 175 Auto-ranging multi-meter, used as an ammeter.4. The diffraction tube (Tel-Atomic)
The 6.3 volt AC supply will provide current to a filament that heats the cathode of thediffraction tube. Once the cathode temperature is high enough, it will eject electrons. Theseelectrons will be controlled the 60 volt DC power supply, connected to the cathode. This isdone in such a way that electrons with insufficient energy to escape this voltage differenceare returned to the cathode. Such a measure controls the number of electrons that will formthe beam; too many in the beam will burn a hole in the graphite target. The total currentmust be kept below 100µA to prevent damage to the target.The electrons will be accelerated by the 0− 5000 volt DC power supply, connected betweenthe cathod can and anode, the actual voltage value determines the electron momentum ac-cording to eV = mc2(γ − 1) ≈ 1
2mp2
The graphite target is powdered graphite bonded to a nickel mesh. The layer is extremelythin. Powder is used so that many different scattering planes are presented to the beam,resulting in a circular ring of scattered electrons striking a fluorescent screen in the tube.
53
r
Dx
L
Anode
Graphite target
Cathode
Cathode can
I
Ammeter0-5000 VDC
0-60V DC
Electronbeam
6.3 VAC
θ
6.2 Background
Graphite is a carbon allotrope consisting of very large, parallel planes of benzene rings, withan interplanar separation of about 3.5 A, with inter-atomic spacings d1. This distance isabout 1.40 A, which is intermediate between a single carbon-carbon bond of 1.54 A and adouble bond of 1.33 A.
3.5 A
There are planes separated by about 7A that perfectly superimpose, separated by planeshalfway in between. These half-way planes have Carbon atoms in locations that superimpose
54
on the centers of the hexagons in the planes above and below. The stereogram below mayhelp you see this.
There is an upper (green) plane, a lower (blue), and an interleaving (red) plane in the figure,each separated by 3.55A with intercarbon spacings in each plane of 1.4A to scale. See Solid
State Physics, by Ashcroft and Mermin, p. 304 for details.The spacings between atoms do not serve aperatures through which the electrons diffract;the spacing is too big for that. What we see instead is reflection of the electron beam fromconsecutive planes of atoms. The angle of incidence of the beam with respect to the planenormals must be just right in order that beams reflected from consecutive planes interfereconstructively by time they reach the detector (glass shell of the diffraction tube). Thiscondition is called the Bragg condition From the figure we can see that when the incomingbeam arrives at the first plane of atoms, some of the beam is reflected, some transmitted toreflect from the next plane. It is these two reflected beams that interfere. The optical pathdifference between the two beams must be an integer number of wavelengths in order thatthe beams constructively interfere when they reach a detector, this path condition is
2h− s = mλ
according to the figure. From the geometry alone we see that
s = 2x sinα, x = h sinα, d = h cosα
and the inter-planar spacing isd = d1 sinα
55
d 1
d
α
α α
α α
xh
s
to screen
incominge
θ=π−2α
from which we obtain
mλ = 2h− 2x sinα = 2h(1− sin2 α) = 2d cosα = 2d1 sinα cosα
ormλ = d1 sin 2α = d1 sin θ
in which d1 is the spacing between atoms, and θ is the scattering angle, the angle betweenincoming and outgoing beams. We will look for the m = 1 bright interference band, theprimary.
The voltage used to accelerate the electrons gives the kinetic energies that are nowhere nearrelativistic, and so we can use
p = mv, K =1
2mv2 =
1
2mp2
and the de’Broglie condition
λ =h
p
together with conservation of energy
eV =1
2mv2
to obtain the electron wavelength in terms of the accelerating voltage
56
λ =h√
2meV=
1.2268× 10−9mV olt12√
V
we find that after using the Bragg condition
1√V
=
√
2me
h2d1 sin θ
is the relation between accelerating voltage and the scattering angle for the electron beam.If we plot 1√
Vversus sin θ, the slope will give us d1, the inter-atomic spacing.
Examine the picture of the apparatus. We can measure the scattering angle θ by measuringthe radius of the ring that the scattered electrons leave on the fluorescent tube. Notice
tan θ =D2
L− r + x
and
x =
√
r2 − (D
2)2
or
θ = tan−1(D2
L− r +√
r2 − (D2)2
)
We see that we must produce a plot of sin θ versus 1√V
and run a linear regression on thedata to obtain the slope of the line of best fit in order to get the inter-atomic spacing. Forthis apparatus we find the physical parameters are r = 6.6cm, the radius of curvature of thebulb, and L = 14.0cm, the length of the bulb.
6.3 Data and analysis
Dinner Douter V Icm cm kV µA
2.68 4.59 3.9 40.72.79 4.77 3.4 39.72.93 5.08 3.1 34.03.05 5.31 2.8 50.03.22 5.59 2.5 42.03.44 6.00 2.2 48.63.69 6.51 1.9 52.3
We have recorded the diameters of two rings, inner and outer, and angular displacement ofthe diffraction rings.We process the data by first put this data into a file as follows
57
2.68 4.59 3.9
2.79 4.77 3.4
2.93 5.08 3.1
3.05 5.31 2.8
3.22 5.59 2.5
3.44 6.00 2.2
3.69 6.51 1.9
called “data” and run it through several Awk scripts to compute the average θ,
awk ’ print sin(atan2(($1)/2.0, 7.4+sqrt(43.56+($1)*($1)/4.0)))," ",
1/sqrt(1000.0*$3)’ data
which results in the data for the inner ring
0.0943793 0.0160128
0.0981397 0.0171499
0.102907 0.0179605
0.106975 0.0188982
0.112709 0.02
0.120078 0.0213201
0.128375 0.0229416
which we can LaTeX format as a table
Inner ringsin(θ) 1√
V
0.0943793 0.01601280.0981397 0.01714990.102907 0.01796050.106975 0.01889820.112709 0.020.120078 0.02132010.128375 0.0229416
The graph of this is shown below, and linear regression
58
Inner ring
0.09 0.10 0.11 0.12 0.13 0.140.016
0.017
0.018
0.019
0.020
0.021
0.022
0.023
sin(θ)
1/√V
results in a line of best fit with 1√V
= −0.002513+ 0.0.198905 sin θ , χ2 = 0.000007 which isa very good fit. From this we find that
d1 =0.198905
8.14367× 108= 0.0244× 10−8m = 2.44Angstroms
with an accepted inter-carbon in-plane separation of 1.40Angstroms, it is pssible that thisring represents scattering off of the large parallel planes of benzene rings, for which our valueis in error by 31 %.It is more likely that we are measuring twice the 1.21 A separation illustrated below
1.4
1.21
for which our results are in excellent agreement of 0.8 %.
For the second ring we run
59
awk ’ print sin(atan2(($2)/2.0, 7.4+sqrt(43.56+($2)*($2)/4.0)))," ",
1/sqrt(1000.0*$3)’ data
resulting in the data
Outer ringsin(θ) 1√
V
0.157521 0.01601280.163204 0.01714990.17287 0.01796050.179941 0.01889820.188429 0.020.200617 0.02132010.215368 0.0229416
and run the linear regression for the line of best fit;1√V
= −0.002244 + 0.117366 sin θ with χ2 = 0.000008, resulting in
d2 =0.117366
8.14367× 108= 0.0144× 10−8m = 1.44Angstroms
which is evidently scattering off of a different set of planes, and is a good value for thein-plane inter-carbon spacing of 1.4 A.
Outer ring
0.14 0.16 0.18 0.20 0.220.016
0.017
0.018
0.019
0.020
0.021
0.022
0.023
sin(θ)
1/√V
60
6.4 Crystal structure determination by scattering
In a crystal, the spatial structure is invariant under translations by vectors
T = na +mb + pc, m, n, p integers
forming a periodic structure falling into 14 types called Bravais lattices. The notation forthis is usually an ordered triple with no parenthesis mnp or commas. The atoms in thecrystal form a periodic structure that superimoses upon itself if it is moved by any of thesetranslation vectors.
14 Bravais latticesSystem Number of types Symbol CharacteristicsTriclinic 1 P a 6= b 6= c, α 6= β 6= γ
Monoclinic 2 P, C a 6= b 6= c, α = 90 = γ 6= β
Orthorhombic 4 P, C, I, F a 6= b 6= c, α = 90 = γ = β
Tetragonal 2 P, I a = b 6= c, α = 90 = γ = β
Cubic 3 P=sc, I=bcc, F=fcc a = b = c, α = 90 = γ = β
Trigonal 1 R, I a = b = c, α = γ = β < 120, 6= 90
Hexagonal 1 P a = b 6= c, α = 90 = β, γ = 120
a
b
c
α
β
γ
For the important example of a body-centered cubic structure, the translation vectors T areillustrated below. There are two cells in these figures, the conventional cell, which is therectangular cubic structure,
61
a
cb
a
cb
and the primitive cell which is the parallelopiped defined by the three lattice vectors a, cand c, which has volume |(a×b) · c|. A stereogram of the primitive cell for a body-centeredcubic lattice (in red) is illustrated below.
The Wigner-Seitz primitive cell is gotten by picking a lattice site, drawing a connectingline to each nearby lattice site, and bisecting each of these lines with a plane. The smallestvolume contained by this structure is the primitive cell.
A stereogram of the primitive cell for a face-centered cubic lattice (in red) is illustratedbelow.
If you look at these pairs correctly, you will be able to see the three dimensional structureof the crystal. Be careful not to break your brain.Solid state physicists propagate some confusion by reporting atomic positions for cubic lat-tices using the conventional cell coordinates.Aluminum is a good example of a face-centered-cubic structure with the length of a side ofits conventional cell being 4.04 A and an atomic nearest neighbor distance of 2.86 A.
62
Once primitive cells have been established, Atoms can be placed at various points withinthem, with the convention that the center of a primitive cell is at 1
212
12. A common error that
people make is to think that the atoms of the lattice are located on the vectors T, when theatoms are merely separated by these vectors. For example, in a NaCl crystal, which is fcc-cubic, each primitive cell contains four Sodium and four Chlorine atoms, Sodiums being at000, 1
2120, 1
201
2, 01
212
and Chlorines at 12
12
12, 001
2, 01
20, 1
200 in conventional cell (Cartesian)
coordinates. These atoms and there positions in the conventional and primitive cells areillustrated in the stereogram below.
The length of the side of the conventional cell in the NaCl structure is 5.63 A.Various types of waves incident on crystals interact with the atoms at lattice points, andwith planes upon which families of atoms lie. The Miller indices of planes of atoms appearin the scattering formulas for these waves. Consider for example families of planes of atomsparallel to a plane defined by three atoms that lie on it, for example 3a, 2b, and 3c. TheMiller indices of the plane are gotten by taking the reciprocals of these integers, namely13, 1
2, 1
3and constructing the smallest set of integers in the same ratio, for our example this
is (232). This triplet is called the set of Miller indices. In applications of these rules, a planethat intersects a crystal axis at infinity has the corresponding index equal to zero.
Consider waves of k-vector k = (0, 0, 2πλ
) incident on an array of atoms at positions T =ma + nb + pc. The waves will arrive with a phase of
ψ(n,m, p) = eik·T
relative to the wave arriving at the atom at the origin. Consider spherical waves arrivingat a distant observation point in direction n from the atom at the origin, and a distanceR away. The phase of the wave scattering from the atom at T, arriving at the observationpoint is
Ψ(n,m, p) = eik·T+ikrn,m,p
in which rn,m,p is the vector between the atom at T and the distant point, so that
T + rn,m,p = Rn
Thenr2n,m,p = (T−Rn) · (T−Rn) ≈ R2 − 2Rn ·T, rn,m,p ≈ R − n ·T
63
and our single arriving wave has phase
Ψ(n,m, p) = ei(k−kn)·T+ikR
We now add up all of the scattered waves coming from all of the atoms in the lattice fromwhich the waves scatter;
Ψ =∑
m,n,p
ei(k−kn)·(ma+nb+pc)+ikR
= (∑
m
eim(k−kn)·a+ikR)(∑
n
ein(k−kn)·b+ikR)(∑
p
eip(k−kn)·c+ikR)
= eiNkR (∑
m
eim(k−kn)·a)(∑
n
ein(k−kn)·b)(∑
p
eip(k−kn)·c)
Each sum appearing in this formula is in fact a geometrical series, very reminiscent of theamplitudes for multiple slit interference, or diffraction gratings. If we use the sum
M∑
p=0
eipA =1− ei(M+1)A
1− eiA, |
M∑
p=0
eipA| = sin(M+12A)
sin(A2)
we discover a wave amplitude of
|Ψ| = sin(M+12
(k− kn) · a)
sin( (k−kn)·a2
)
sin(M+12
(k− kn) · b)
sin( (k−kn)·b2
)
sin(M+12
(k− kn) · c)sin( (k−kn)·c
2)
in which M is some large integer. As M becomes very large, in other words as the size of thecrystal that we are scattering off of increases, this amplitude becomes very sharply peakedaround its central maximum, being just a product of three diifraction grating amplitudes.The maximum occurs for n directions for which all three factors in the product are one;
(k− kn) · a = 2πma, (k− kn) · b = 2πmb, (k− kn) · c = 2πmc
in which ma, mb, mc are integers. These are the Laue conditions, and in fact are equivalentto the Bragg condition.
6.5 Simulation of scattering
A typical crystal diffraction pattern for waves scattering off of a single crystal has brightspots in directions in which these three conditions are met. We can easily simulate this. Forexample, the (photographic negative) of the diffraction pattern for scattering from a 9×9×9hexagonal, close packed crystal looks like the following
64
In polycrystalline scattering, like off of a powder consisting of many single crystals witharbitrary orientations, the diffraction pattern is similar to that above, but is gotten from thespot pattern by rotation to form concentric rings.Below you will find a simple C program for simulating this experiment and testing the Laueconditions. You may make changes in it, to output the angles θ and φ of the bright spots,so that you can verify the conditions by hand.
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<plot.h>
#define PI 3.1415926
__complex__ double phase,eye;
__complex__ double cexp(__complex__ double num);
double modulus(__complex__ double num);
int m,n,p,M,N,pl_handle,num;
double a[3],b[3],c[3];
double r[3],Q,R,S;
double k[3],kp[3],spot[1000][3];
double phi,theta,x,y,dx,dy,amp,R,lambda;
main()
/* lattice vector components */
a[0]=2.0,a[1]=0.0,a[2]=0.0;
b[1]=-sqrt(3.0),b[0]=1.0,b[2]=0.0;
c[2]=2.0,c[0]=0.0,c[1]=0.0;
eye=0.0+1.0i;
dx=0.05;
dy=0.05;
/* x-ray wavelength */
lambda=0.05;
k[2]=2.0*PI/lambda;k[0]=0.0;k[1]=0.0;
65
pl_handle = pl_newpl ("ps", stdin, stdout, stderr);
pl_selectpl (pl_handle);
if (pl_openpl () < 0) /* open Plotter */
fprintf (stderr, "Couldn’t open Plotter\n");
return 1;
pl_fspace (-4.0, -4.0,4.0, 4.0);
pl_filltype(1);
pl_fillcolorname("white");
/* pl_fbox(-4.0, -4.0,4.0, 4.0); */
pl_flinewidth (0.005);
for(M=-40;M<40;M++)
for(N=-40;N<40;N++)
/* using over 10,000 points on viewing screen */
R=sqrt(((double)N*dx)*((double)N*dx)+((double)M*dy)*((double)M*dy));
theta=atan2(R,30.0);
phi=atan2((double)M*dy,(double)N*dx);
/* this is k’- the elastically scattered wavevectors */
kp[0]=(2.0*PI/lambda)*cos(phi)*sin(theta);
kp[1]=(2.0*PI/lambda)*sin(phi)*sin(theta);
kp[2]=(2.0*PI/lambda)*cos(theta);
phase=0.0+0.0i;
/* lattice will be 9x9x9, pretty big */
for(m=-4;m<=4;m++)
for(n=-4;n<=4;n++)
for(p=-4;p<=4;p++)
/* generate a lattice scattering site */
r[0]=(double)m*a[0]+(double)n*b[0]+(double)p*c[0];
r[1]=(double)m*a[1]+(double)n*b[1]+(double)p*c[1];
r[2]=(double)m*a[2]+(double)n*b[2]+(double)p*c[2];
phase=phase+cexp(eye*(r[0]*(k[0]-kp[0])+r[1]*(k[1]-kp[1])+\
r[2]*(k[2]-kp[2])));
amp=modulus(phase);
/*printf("%f\t%f\t%f\n",(double)N*dx, (double)M*dy, amp);*/
num=(int)(65535.0*(amp/729.0));
if(num>4000)
pl_color(num,num,num);
/* output a spot (hopefully) corresp. to reciprocal wavevector */
pl_fbox((double)N*dx-dx/2.0,(double)M*dy-dy/2.0,(double)N*dx+dx/2.0,(double)M*dy+dy/2.0);
/* check for laue conditions */
Q=(a[0]*(k[0]-kp[0])+a[1]*(k[1]-kp[1])+a[2]*(k[2]-kp[2]))/(2.0*PI);
R=(b[0]*(k[0]-kp[0])+b[1]*(k[1]-kp[1])+b[2]*(k[2]-kp[2]))/(2.0*PI);
S=(c[0]*(k[0]-kp[0])+c[1]*(k[1]-kp[1])+c[2]*(k[2]-kp[2]))/(2.0*PI);
66
/* output Q=m_a, R=m_b, S=m_c for brightest spots */
printf("%f & %f & %f & %f\\\\\n",100.0*(float)num/(float)65535.0,Q,R,S);
pl_closepl ();
pl_selectpl (0);
pl_deletepl (pl_handle);
__complex__ double cexp(__complex__ double num)
__complex__ double tmp;
__real__ tmp=exp(__real__ num)*cos(__imag__ num);
__imag__ tmp=exp(__real__ num)*sin(__imag__ num);
return(tmp);
double modulus(__complex__ double num)
double tmpx,tmpy;
tmpx=__real__ num;
tmpy=__imag__ num;
return(sqrt(tmpx*tmpx+tmpy*tmpy));
The centers of the very brightest spots are organized in the table below, and the “integers”for the Laue conditions are displayed. You can see that even for a mere 729 atoms, thescattering maxima obey the Laue conditions quite well.
67
Laue bright spots% brightness ma mb mc
43.862058 0.998751 -0.999860 0.04996243.837644 -0.998751 -1.998611 0.04996257.804227 -0.000000 -1.038763 0.01798884.116884 -0.000000 -0.981102 0.01604682.714580 0.999575 -0.019607 0.01698983.109789 -0.999575 -1.019182 0.01698959.916075 0.999599 0.038104 0.01604659.301137 -0.999599 -0.961494 0.01604643.884947 1.997505 0.998752 0.049906100.000000 0.000000 0.000000 0.00000043.884947 -1.997505 -0.998752 0.04990659.301137 0.999599 0.961494 0.01604659.916075 -0.999599 -0.038104 0.01604683.109789 0.999575 1.019182 0.01698982.714580 -0.999575 0.019607 0.01698984.116884 -0.000000 0.981102 0.01604657.804227 -0.000000 1.038763 0.01798843.837644 0.998751 1.998611 0.04996243.862058 -0.998751 0.999860 0.049962
Read through the program and determine how far the crystal sample is from the screen onwhich these spots are cast, and find the wavelength. Use the data to actually determinethe lattice spacings. This might give you some insights into your experiment in electrondiffraction.
68
7 The Photoelectric Effect
In this experiment, the interpretation of which Einstein received the Nobel prize, we measurePlanck’s constant and verify the particulate nature of light. It provides an excellent contrastto the electron diffraction experiment that illustrates the wave nature of matter.
7.1 Apparatus
1. A light source. Welch Scientific model 36 Universal light source.2. Planck constant apparatus; Sargent-Welch Scientific, catalog number 2120.
3. 10 cm and 2.5 cm focusing lenses.4. Voltmeter. Keithley 610c solid state electro-meter.5. Ammeter. Keithley 485 auto-ranging picoammeter.6. 3 volt power supply. 2 1.5 volt size D batteries in series.
The 3 volt power source provides a fixed anode/cathode voltage difference and will be usedto measure the maximum energy of the ejected photo-electrons. There are a pair of variableresistors R1 and R2 built into the Planck apparatus. They are used to precisely control theanode/cathode voltage. A voltmeter is used to exactly measure this anode/cathode voltage.The role of the ammeter is to measure the current of the photo-electrons. When the currentis zero, no electrons have sufficient kinetic energy to reach the cathode.Filters built into the apparatus restrict the frequencies of light entering the device to a verynarrow bandwidth. We really only want light of one frequency entering the device, to besure that we know precisely which wavelengths cause photo-emission. The lense is used toeliminate other light sources andto direct the beam from the universal light source. Thesource itself is as close to white light as money (a little, not a lot) can buy. It provides thephotons that cause photo-emission of electrons from the target metal (anode).
69
3 V DC
V
Voltmeter
I
Ammeter
Cathode
Electric field
White light
LenseChangeablefilter
Anodeelectron beam
R1 R2
7.2 Background
In the Photoelectric effect, light incident on the anode can be absorbed by electrons , andif they absorb enough energy, the electrons can be liberated from their potential wells andescape the metal. The kinetic energy of the electron plus the energy required to liberatethem from the anode must equal the energy of the photon, here assumed to be a discreteparticle with energy related to its frequency
K + φ = hν
φ is called the work function, and naturally varies from metal to metal. In terms of lightwavelength this is
K + φ =hc
λ
The kinetic energy K can be measured by applying sufficient voltage to stop the electron
K = eVstop
and we find that
Vstop =h
eν − φ
e
70
stopping voltage versus light frequency is a straight line whose slope gives the value ofPlanck’s constant and intercept determines the work function of the metal.
7.3 Procedure and data
We will apply various voltages to the anode/cathode and monitor the current for severallight wavelengths.
577 nm filter 546 nm filter 435.8 nm filterV I I I
volt µA µA µA
0.050 .3475 .6206 1.65680.150 .2009 .3882 1.2375.250 .0971 .2070 .8615.350 .0373 .0928 .5461.450 .0098 .0329 .3227.550 -.0001 .0066 .1768.650 -.0027 -.0018 .0898.750 -.0034 -.0045 .0404.850 -.0036 -.0052 .0127.950 -.0036 -.0055 -.00161.05 -.0037 -.0056 -.00901.15 -.0038 -.0057 -.01191.25 -.0039 -.0058 -.01301.35 -.0040 -.0058 -.01341.45 -.0041 -.0059 -.01361.55 -.0041 - -.01381.65 -.0042 -.0060 -.01401.75 -.0042 - -.01412.00 - -.0061 -.01432.10 -.0043 - -.01432.50 -.0043 -.0062 -.01452.75 -.0043 - -
We now graph this data ( voltage versus current, 435.8 nm, 546 nm, 577 nm respectively)and determine the stopping voltage by looking for the place where the data curves start todeviate from the straight line portion, since no current passes due to flowing electrons oncethis point is reached.
71
0.0 0.5 1.0 1.5 2.0 2.5−0.5
0.0
0.5
1.0
1.5
2.0
0.0 0.5 1.0 1.5 2.0 2.5−0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.0 0.5 1.0 1.5 2.0 2.5 3.0−0.05
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
By doing a spline fit we find that for 577 nm light ( ν = 5.1957× 1014Hz)
Vstop = 0.7167± 0.0866 volts
for the 546 nm ( ν = 5.4907× 1014Hz) data we obtain
Vstop = .8667± 0.0866 volts
and for the 435.8 nm (ν = 6.8791× 1014Hz) data
Vstop = 1.33± 0.0866 volts
This we now graph, stopping voltage versus frequency or else do a linear regression on thedata, to get the slope of the best fit line
72
4.5 5.0 5.5 6.0 6.5 7.00.4
0.6
0.8
1.0
1.2
1.4
The linear regression gives us a slope of
0.356822± 0.032069volt
1014Hz=h
e
from which we geth = 5.17163± 1.0913× 10−34J · sec
which is a rather good result , close to the accepted value of 6.6262× 10−34J · sec
8 Measuring Planck’s constant (solid state version)
In this version of the experiment you will use the Klinger LED apparatus, containing a se-lection of six LED (light emitting diodes) in series with a 100 Ω resistor. Hopefully you willbe able to obtain five or even six data points for the Planck line.In addition you will need a 2− 12V AC variac and two voltmeters (DMMs).
8.1 Theory
When free atoms are brought together to form a periodic-lattice solid, the multiply degen-erate energy levels of the free atoms blend into energy bands, near continuous collections ofelectron levels. The valence electrons of the seperated atoms tend to form into one band,the valence band, in the solid, and bonding electrons often compose a seperate higher en-ergy band, the conduction band. The presence of electrons or other charge carriers in theconduction band determine electrical conductivity of the solid, and there is usually a gapbetween the valence and conduction bands.
73
Seperate atoms Solid state
Conduction band
Valence band
Eg
We know that an n-type semiconductor has electrons in its conduction band, and a p-typehas some holes in the valence band. Before an n-type and p-type are joined to form a pn-junction, the electron energy diagrams for the two look like the following;
E
Distance
Valence
Conduction
p−type n−type
The gray areas represent filled electronic band states. Once the two materials are physicallyjoined, there are electron and hole concentration gradients at the junction; on the n-side,there is a higher concentration of electrons than on the p-side, and on the p-side the holeconcentration exceeds that of the n-side. Concentration gradients drive diffusion, and soelectrons will diffuse from the n-type into the p-type, and holes diffuse from p-type to n-type. Holes and electrons then “annihilate” one another, this forms a thin layer around thejunction in which there will be excesses of the appropriate donor or acceptor ions.
74
E
Distance
Valence
Conduction
p−type n−type
− +
This thin layer is called the space-charge region, and on the n-side of it there will bean imbalance of acceptor ions, on the p-side an imbalance of electron donor ions, and thiscreates an electric field and potential in the space-charge region. This is called the built-infield. The field exerts forces on electrons and holes that opposes the diffusion, allowing anequilibrium the result.Examine the electrostatic potential energy of electrons on both sides of the junction, sinceEbuilt−in = −dV
dxand outside of the space-charge region the two materials are electrically
neutral, we see that a solution to the Poisson equation is a constant potential Vp in thep-type material, a constant Vn in the n-type, and the ramping potential due to the built-infield connects the two in the space-charge region.
V(x)
Vp
Vn
Ebuiltin
EF
p−type n−type
E
Conduction band
Valence band
Multiplication of this potential by the electron charge (negative) results in the electronicenergy structure illustrated above for a pn-junction at equilibrium, here illustrated with aFermi level indicated by a dotted line.
75
Consider the energy level scheme of a pn-junction forming the boundary between an n-typesolid with a partially filled conduction band (filled electron states) up to some Fermi level(indicated on the right), and a p-type solid with holes acting as charge carriers. On thep-side, empty electron states (holes) in the valence band are not shaded. The gap is drawnon the p-side for convenience; there are no empty hole states in the conduction band onthe p-side, and the filled hole states fall short of the valence-to-conduction band gap on thep-side.If the junction is forward-biased, the energies of the n-type bands will be raised to a pointsuch that the barrier presented to the flow of electrons from the n-side is dramatically re-duced, and electrons can migrate into the lower levels of the conduction band on the p-sidefrom the n-side, where they combine with holes.
EF
p−type n−type
E
Filled e− states
(n−conduction band)
Empty states (holes)
not shaded
(p−valence band)
This results in the creation of a photon whose energy is the band-gap energy(between valenceand conduction bands).
76
p−type n−type
E
hν=Egap
|e| Vmin
If the band-gap is at least Eg ≥ 1.5 eV , the emitted photon will be visible. In additionholes can migrate from the p-side to the n-side, combine with electrons and emit the samefrequency light. This process may be a bit easier to understand; the hole migrating to then-type side is a vacancy in the valence band on the n-side, which “annihilates” an electronby providing a vacancy that an electron in the conduction band can fill (to oversimplify thematter somewhat).
p−type n−type
E
hν=Eg
|e| Vmin
In any event, when the applied forward bias is at the threshold for a charge carrier transferfrom one side to another, one can see that we have
|e|Vmin = hω = hν, |e|Vmin ≤ Eg
77
which allows us to measure Planck’s constant and verify the Planck Hypothesis by measuringthe threshold voltage for forward-biased conduction and plotting it versus LED dominantradiated frequency. In the real world the diode emits a variety of frequencies, but the dis-tribution is fairly sharp at a precise value. The wavelengths of the dominant photons areprinted right on the apparatus.
8.2 Procedure
It should be fairly clear what needs to be done; for each diode a plot of current versus voltageacross the diode needs to be produced. The current will be very small, but not zero (due totunneling) until the voltage drop reaches the required Vmin for forward conduction, at whichpoint it will climb rapidly.
V
DMM
100 Ω mA
Do this for each diode, and use the computer program for fitting data to polynomial curvesto fit that portion of your data for which I > 0 to a high order polynomial such asI = a4V
4 + a3V3 + · · ·a1V + a0, and use the method of bisection program to find the
roots or zeros of this function. This will give you Vmin in each case. Both programs can beobtained from the course website.
Example Suppose that I took the (ficticious) current versus voltage data below.
78
0.0 0.5 1.0 1.5 2.00
1
2
3
4
5
6
7
V
I(m
A)
I have divided the data into two sets, that with current equal to zero (squares) and thatwith current greater than zero (circles). I now run a least-squares polynomial fit on thecircle-data, obtaining
I(V ) = −0.7Amp+ 0.3Amp
V olt2V 2 + 0.5
Amp
V olt4V 4
as the best fitting quartic, the fifth and sixth order fits being worse by a Chi-squared crite-rion. I obtain Vmin by the method of bisection or Newton’s method, beginning with a guessnear 1.0V .
8.2.1 Finding roots by bisection
To apply the bisection method, pick two points, x1 and x2. If there is a root between x1 andx2, then
f(x1)f(x2) < 0
otherwisef(x1)f(x2) > 0
The first step in the algorithm is to test which case prevails.
If there is a root between x1 and x2, bisect the segment and let z = x1+x2
2. Now test if there
is a root between x1 and z. If there is , reset x2 = z and iterate. If there was not, then there
is a root between z and x2 and so we reset x1 = z and iterate. We continue to do this until
we have narrowed down the interval in which there is a root to whatever accuracy we desire.
The simple C + + program below accomplishes this.
79
#include<iostream.h>
#include<math.h>
float f(float x); /* define your function in this routine */
main()
float x0,x1;
cout<<" Input x0 and x1, two points around a root"<<"\n";
cin>>x0>>x1;
for(int n=0;n<20;n++)
if (f(x0)*f((x0+x1)/2.0)<0.0)
x1=(x0+x1)/2.0;
else x0=(x0+x1)/2.0;
float root=(x0+x1)/2.0;
cout<<"the root is = "<<root<<"\n";
float f(float x)
return(x*x-3.0*x+1.0);
Once you have the least squares polynomial fit, you can create a subroutine that com-putes the polynomial, and point the function pointer to your routine.
8.2.2 Newton’s method
The Method of Newton is another iterative scheme for finding or refining roots that canbe potentially unstable, and so should be used with caution. If
f(xn+1) ≈ f(xn) + (xn+1 − xn)f ′(xn) ≈ 0
then
xn+1 ≈ xn −f(xn)
f ′(xn)
so if xn is close to a root, xn+1 should be closer. The disadvantage, other than stability, isthat you need the derivative function as well as the function itself in order to evaluate thisexpression. If you have both functions, this can be used to polish or refine an estimate of aroot gotten by bisection.
80
One distinct advantage of the Newton method over bisction is that since it is non-geometrical,
it applies equally well to real or complex roots. One needs to make sure that if a complex
root is sought, a complex initial guess is used, otherwise polynomials with real coefficients
are always real-valued. For example
/* Newton-Rafeson computation of roots of cubic */
/* for complex root, make sure x0 (first guess) */
/* is a complex number */
#include<stdio.h>
int n;
__complex__ double x0,x1,y;
__complex__ double f(__complex__ double x);
__complex__ double df(__complex__ double (*pf)(__complex__ double x), \
__complex__ double x); /* derivative routine accepts pointer to function */
double modulus(__complex__ double z);
main()
x0=1.0+0.9i;
do
y=x0; /* test for precision */
x1=x0-f(x0)/df(f,x0); /* pass derivative a pointer to f */
printf("%f+%fi\n",__real__ x0,__imag__ x0);
x0=x1;while(modulus(y-x0)>=1.0e-15);
return;
__complex__ double f(__complex__ double x)
return(x*x*x+1.0);
__complex__ double df(__complex__ double (*pf)(__complex__ double x), \
__complex__ double x)
__complex__ double del,y;
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del=0.1;
y=(8.0*(*pf)(x+del)-8.0*(*pf)(x-del)-(*pf)(x+2.0*del)- \
(*pf)(x-2.0*del))/(12.0*del);
return(y);
double modulus(__complex__ double z)
double re,im;
re=__real__ z;
im=__imag__ z;
return((re*re+im*im));
At any rate these programs give you at least two options for computing Vmin fromI(Vmin) = 0, and you can devise other methods as well.
Your lab report will contain the current versus diode voltage drop for all six diodes in tabularform, graphs of all six sets of data, the set of points used for the polynomial fit must beclearly marked since I will perform the fit to check your work, and the actual poly-nomial, fourth or higher order. You will also report a table of Vmin versus diode wavelengthand frequency or each diode, and finally produce a graph of Vmin versus f = ν, along witha least squares fit to determine Planck’s constant from the slope, and an error estimate forit. You must compare your value of Planck’s constant with the accepted value.All of these programs compile on any Linux or FreeBSD computer, and on CYGWIN orMINGW32 as well. The lab computers are equipped with CYGWIN, and LaTeX for yourconvenience.
82
9 Helium Ionization Potential; The Franck-Hertz Ex-
periment
How do electrons interact with ordinary atomic matter? In 1914 J.Franck and G. Hertzperformed an experiment in which a beam of electrons is passed through a gas of atoms.Some of the electrons will lose energy to the atoms (originally a gas of mercury vapor) . Theelectrons are first accelerated by a voltage Vc between the apparatus cathode and a grid, andalso by a decelerating voltage Va between the anode and grid. Electrons should arrive atthe anode with energy e(Vc− Va), unless of course they interact with the gas atoms. In thiscase they arrive at the anode with a final energy Ekin = e(Vc−Va)−∆E. As anode current,which measures the energy of electrons arriving at the anode, is monitored as a function ofVc, we see fluctuations in the current at precise intervals.
Hg sample
0 5 10 15 20 250
5
10
15
20
25
V, volts
I mill
iam
ps
As the accelerating voltage is increased, electrons arrive at the collector ring in the apparatusat an increasing rate, and so we expect the collector ring current to increase with acceleratingvoltage. This does occur, until the incoming electrons have enough kinetic energy to excitethe Mercury atom in the tube. Then an inelastic collision takes place, the incoming electronvirtually stops dead, the Mercury atom is excited, and the ring current drops since electronsare not reaching it.This indicates that there is a threshold energy for electron/atom interactions, and that thethresholds occur at energies regular intervals of 4.9eV for mercury. What we are seeing is
83
the projectile electron giving up energy to promote bound electrons in the mercury atom tohigher atomic energy levels. This is reinforced by the fact that each 4.9eV energy loss isaccompanied by photon emissions from the mercury, of light with a wavelength correspond-ing to this energy difference. This is very strong evidence for the discrete nature of boundelectron energy states.
We perform a variation on the classic experiment here at Parkside. Instead of mercury,we excite a gas of helium atoms with the electron beam. This particular apparatus may bephased out, since it is a bitch to set up. A short description with data for a newer apparatusfollows this section.
9.1 The old version of the experiment; apparatus
1. 1 MegaOhm resistor. Cornell-Dublier electronics decade resistor model RDC.2. 1.5 V power supply. Single 1.5 volt size D battery.3. 20 V power supply. Heathkit IP-2718.4. Signal generator. Wavetek model 182A function generator.5. Ammeter. Keithley 175 auto-ranging multi-meter.6. Voltmeter. Keithley 178 digital multi-meter.7. 12 volt power supply. 12 volt automotive battery, Farm’n Fleet model UL-300CA.8. Rheostat. Cenco 22 ohm, 4.4 amp , catalog number 8291013.9. Franck-Hertz tube. Tektronics Franck-Hertz apparatus.
84
These are assembled as shown below.
A
V
-+
- +
Ammeter
Rheostat
Voltmeter
12 V Powersupply
6 V peak-to-peaksignal generator
0-20 V power supply
collecting ring
anode
1 M
1.5 V DC Oscilloscope
Filament
We briefly describe the role each item plays in the experiment.Oscilloscope monitors the voltage across the 1MΩ resistor, through which runs the currentof electrons that arrive at the collector ring. Increases in the electrons making it to the ringincrease the current, and the voltage across the resistor. The scope also monitors timesat which the voltage reaches a value that indicates absorptions by the helium gas. This issignaled by sudden drops of voltage across the resistor, decreases in electron flow translateinto drops in beam kinetic energy due to energy absorption by the gas.The resistor is in the circuit simply to aid in measuring the voltage or current from thecollection ring.1.5 volt battery is used to raise the collection ring voltage slightly above the anode voltage.Electrons that lose energy in collisions with helium atoms will be moving slowly enough tobe collected.20 volt power supply is used raise the saw-tooth voltage from the signal generator to sucha point that the peak to peak value of the total voltage input spans the full energy range forexcitation of the helium, from 20 to 26 volts.The signal generator provides a predictable voltage that increases at a determined rate.Rather than performing the entire experiment with several cathode voltages, we perform itwith an AC sawtooth voltage that sweeps through the entire excitation range periodically.The ammeter monitors the filament current, which cannot exceed 1.5 amps without beingdamaged.The voltmeter monitors voltage across the filament, which needs to be kept below 5 volts
85
in order not to damage the filament. Another safety precaution.12 volt power supply provides a constant DC voltage to the filament. A battery is moreconstant than an electronic power supply, no ripples or fluctuations. This produces a con-stant production of thermionic electrons.The rheostat limits the flow of current to the filament, again to prevent burn out.The Franck-Hertz tube contains a filament from which electrons are expelled by thermionicemission. These will be accelerated and will collide with helium in the tube. Varying thevoltage on the anode creates the accelerating potential difference for the emitted electrons,which are collected by the collector ring.
9.2 Data and analysis
Once the apparatus is assembled and all voltages are in the required range, we can take data.We adjust the 20V power voltage until three excitation peaks are visible on the oscilloscope,and the time of the peaks are recorded in milliseconds
trial 1 trial 2 trial 3ms ms ms
E1 1.9± 0.4 1.9± .4 1.8± .4E2 3.9± 0.4 3.9± 0.4 3.8± 0.4E3 6.1± 0.4 6.2± 0.4 6.0± 0.4
In order to transform these times into voltages, we need to know the slope of the signalgenerator waveform
∆V = 6.0± 0.1 volts
∆t = 3.8± 0.1ms times 2
since the scale on the scope has 2 ms per division. This gives us a slope of
0.789± 0.025V
ms
and we can compute the energies of the excitations
E1 = e(0.789V
ms)(1.87ms) = 1.48eV
E2 = e(0.789V
ms)(3.87ms) = 3.05eV
E3 = e(0.789V
ms)(6.1ms) = 4.81eV
and we can perform standard Gaussian error analysis on our data to arrive at
E1 = 1.5± 0.6eV, E2 = 3.05± 0.55eV, E3 = 4.81± 0.57eV
which agree well with 1.1, 3.1, 4.8 eV respectively, the commonly accepted values.
86
9.3 Second version
This version uses the TEL 2533.01 Helium Hertz critical potential tube, or the This versionuses the TEL 2533.02 Neon tube, and a chart recorder. Recommended as well are the TEL2800 LV and TEL 2801 HV DigiRamp power supplies, and a picommater.The tube containing the gas sample has a weakly conducting inner coating connected to theanode, otherwise it functions pretty much the same as in the previous description; acceleratedelectrons undergo inelastic collisions with gas atoms once they reach sufficient energy to causethe atoms to undergo an atomic transition, otherwise the collisions are elastic. As the beamundergoes further acceleration by the field in the tube, it may reach another critical valueand reveal another atomic energy level. For the Neon tube, a typical chart recorder outputlooks like this;
Neon sample
0 10 20 30 40 500
10
20
30
40
50
V, volts
I
recording collector ring current (from electrons that give up all or most energy to exciteatoms) versus accelerating voltage.In this figure the current peaks at 18.30, 21.22, 39.08, and 41.93V , corresponding to Neonatomic absorptions of 18.30, 21.22, 39.08, and 41.93 eV . These compare very favorably withthe published peak voltages of 17.0, 21.5, 34.0, 38.5V . Remember that it is the sudden dropin current at the peak that signals the prpjectile electron energy loss. This energy loss equalsthe difference in two atomic energy levels of the Neon, and the Neon will then spontaneouslydecay by emitting light at these energies. You can see this with a diffraction grating.
87
Examine the first pair of peaks for this run. The first peak drops off rapidly at E = 18.30 eV ,if the atom were to shed this energy as light, it would be at wavelength
1240 eV · nmλ
=hc
λ= E = 18.30 eV
or λ = 67.8nm, which compares well with the known Neon line at λNe = 73.5nm. Thesecond peak at 21.5 eV corresponds to Neon ionization by the incoming electron.Correcting for the contact potential will improve the accuracy. I would suggest determiningwhich spectral line is being excited for each of your peak dropoffs.Neon has spectral lines at λ = 73.5, 73.4, 585.2, 640.2, 671.7nm.
88
10 The Rydberg Constant
This constant determines the spacing between Hydrogenic Bohr energy levels, and was orig-inally determined spectroscopically. In our modern lab experiments course, you will performthe same experiment. This simply requires a diffraction grating, the theory of which we havestudied in Physics 202. The grating will be incorporated into a spectrometer, and we willstudy the measure the Rydberg constant for hydrogen.
10.1 Background
The “Old Quantum Theory” that supplanted classical mechanics as a description of theelectron in an atom was based We now state the previously derived quantization rule as apostulate Postulate W =
∮
p dq = nh where h is Planck’s constant.
Apply this now to the problem of finding the quantum energy levels of the hydrogen atom,with lagrangean
L =m
2(r2 + r2θ2 + r2 sin2 θφ2)− V (r)
The momenta are
pr = mr
pφ = mr2 sin2 θφ
pθ = mr2θ
and the Hamiltonian is then, by eliminating the derivatives
E = H =p2
r
2m+
p2θ
2mr2+
p2φ
2mr2 sin2 θ+ V (r)
Impose the quantization conditions∮
pθdθ = `h
or pθ = `h and similarly pφ = mh and
∮
pr dr =∮
√
2m(E − V (r)− (`h)2
2mr2) dr = nh
Now insert the form of the potential
V (r) =k
r
and note that the contour integral has two poles, one for large r and one for small, whichwe evaluate using the Cauchy theorem
89
∮ dz
z= 2πi
∮
zkdz = 0
,for any k value other than −1. For small r we neglect all but the last term under the radicalto get
∮
√
0− 2m(nh)2
2mr2= 2πi
√
−`2h2 = −`h
For large values of r we instead throw out the last term in the radical and expand using thebinomial theorem, but only one term will have a residue
∮
√
2m(E − k
r)dr =
√2mE
∮
(1− k
2Er+ · · ·)dr = −2πik
√
m
2E
The final result is
∮
prdr = −`h− 2πik
√
m
2E= nh
This can be inverted to get E, recall that E is in fact negative
E =−mk2
2h2(`+ n)2
The interpretation of this result is very simple; there is an energy level degeneracy dueto the fact that several different states with different angular momenta can have the sameenergy. These states differ in their orbital eccentricity.From our classical mechanics experience we know that planetary orbits are elliptical with asemimajor axis a given by
a =k
2|E|and semiminor axis b determined by both energy and angular momentum
b =L
√
2m|E|
such that the orbital eccentricity is
ε =
√
1− b2
a2=
√
1− 2|E|L2
mk2
For the Sommerfeld calculation these become
h` = b√
2m|E|, k
√
m
2|E| = (n+ `)h
90
Multiply these;h`
b√
2m|E|k
√
m
2|E| =h`a
b= (n + `)h
ora
b=n+ `
`=N
`
in which the energy quantum number is
N = n+ `, E =−mk2
2h2N2
This means that our orbits have quantized semimajor and semiminor axes.
aN =k
2|E|N=
h2
mkN2 = a0N
2
and
b =`
NaN = `Na0
in which a0 = 0.53 Angstroms is the Bohr radius. Some of these orbits are illustrated below
N=1, l=1
N=2, l=2
N=2, l=1
N=3, l=3
N=3, l=2
N=3, l=1
Sommerfeld was able to go one step further, by beginning with a relativistically correctexpression for the energy, all of the integrations are pretty much the same, and the endresult illustrates the fine structure splittings in the energies of certain orbits with the sameN but different `, orbits that would be degenerate (have the same energy) without relativity
91
taken into account.The total energy according to relativity is
E =√
c2~p2 +m2c4 − k
r
which we write as
(E +k
r)2 −m2c4 = c2~p2
and we now insert or old expression for the momentum in polar coordinates from our firstSommerfeld computation;
(E +k
r)2 −m2c4 = c2(p2
r +p2
θ
r2)
but again∮
pθ dθ = `h
and since pθ is conserved (constant) this gives us
pθ = `h
again. We conclude that relativistically
pr =1
c
√
(E2 + 2Ek
r+k2
r2)−m2c4 − `2c2h2
r2
and therefore its corresponding action quantization is
∮ 1
c
√
(E2 + 2Ek
r+k2
r2)−m2c4 − `2c2h2
r2dr = nh
This is exactly what we had before, with a few changes. Integrating exactly the same wayas before we obtain
i
√
k2
c2− `2h2 + i
Ek
c√E2 −m2c4
= hh
If we call kZch
= e2
4πε0 h c= α, the fine structure constant, we can rewrite this as
i√Z2α2 − `2 + i
EZα√E2 −m2c4
= n
and we expand in powers of the fine structure constant, which is very small;
√Z2α2 − `2 ≈ i(`− Z2α2
2`+ · · ·)
and so
iEZα√
E2 −m2c4≈ (n+ `)− Z2α2
2`= N − Z2α2
2`
92
or
m2c4 = E2(1 +Z2α2
N2(1− Z2α2
2`N)2
)
This gives us the final relativistically correct energy quantization rule
E =mc2
√
1 + Z2α2
N2(1−Z2α2
2`N)2
≈ mc21
√
1 + Z2α2
N2 + Z4α4
N3`+ · · ·
≈ mc2(1− 1
2(Z2α2
N2+Z4α4
N3`) +
3
8(Z2α2
N2)2 + · · · )
≈ mc2(1− Z2α2
2N2(1 +
Z2α2
N(1
`− 3
4N)) + · · · )
The first term we recognize as the rest energy of the electron, and all of the rest is thequantized electronic energy
EN,` = −mc2Z2α2
2N2− mc2Z4α4
2N3(1
`− 3
4N) + · · ·
This shows that the different ` elliptical orbits allowed for a given N value are not degeneratebut are split by some very small energies. The fine structure constant is
α =e2
4πε0hc= 0.0072974 ≈ 1
137
and this provides us with a very valuable tool for getting an idea as to how large a correctionto the energy each term represents. The basic electron energy is mc2 = 511 keV . The Bohrorbitals represent states whose energies differ from this by α2mc2 = 511 keV
(137)2≈ 0.02723 keV =
27.23eV , and these fine structure corrections are four orders of magnitude smaller still, ormc2α4 = 511 keV
(137)4≈ 1.451× 10−6keV = 1.451× 10−3 eV . A most remarkable fact is that we
know that Bohr-Sommerfeld theory is very flawed, yet this expression for the fine structurespectrum is exactly correct, even when compared to computations performed with therelativistic version of the Schrodinger equation (the Dirac equation), with only the provisothat ` is replaced by `+ 1
2to correct for electron spin, a factor unknown at Sommerfeld’s time.
93
N=1
N=2
N=3
N=4
E
-13.6 eV
-3.38 eV
-1.50 eV
-0.85 eV
Lyman Transitions (UV)
Balmer Transitions (vis)
Lyman Balmer
UV IR
10.2 Atomic Spectra
By hypothesis, Hydrogenic atoms only radiate light when they jump from one Bohr-Sommerfeldstable orbit to one of lower energy. When this occurs, light of frequency f and wavelengthλ is emitted with
hf =hc
λ= EN,` − EM,`′
The spectrum of emitted radiation is determined by the differences in the energy levels.
Example Consider a simple quantized atomic system in which an electron can have fourenergy levels E1 < E2 < E3 < E4 as shown below. There can be six wavelengths of lightemitted when a gas of such atoms is thermally excited.
94
12
3
4
B R
E E
Levels Transitions
Emission Spectrum
These wavelengths are (from most energetic or most blue, to least energetic or reddest)
hf4,1 = E4 − E1, hf4,2 = E4 − E2, hf3,1 = E3 − E1, hf3,2 = E3 − E2
hf4,3 = E4 − E3, hf2,1 = E2 − E1
a better representation of what one would see of this light were passed through either a prismor diffraction grating would be
B REmission Spectrum
Keep in mind that we will see spectral lines because the light from the atoms will be passedthrough a slit, and so the line on the film or grating is an image of the source or slit.This type of emission spectrum is called discrete line, as opposed to a continuous spectrumin which all colors or wavelengths of light are present, as in the spectrum of light emitted by
95
a very hot opaque body or a heated cavity in a conductor (cavity or BlackBody radiationa’la Planck). For light emitted by Hydrogen this expression becomes
hfn,m =13.6 eV
m2− 13.6 eV
n2, λ−1
n,m == R∞(1
m2− 1
n2)
in which R∞ = mcα2
2his called the Rydberg constant
R∞ = 109737.309 cm−1
What we actually see when we analyze light from thermally excited atoms differs from thepredictions of Bohr-Sommerfeld theory only in two details, some of which are beyond theability of that theory to explain;
1. Not all predicted spectral lines (emitted wavelengths) are actually seen.
2. Not all emitted spectral lines have the same intensity or brightness.Basic statistical mechanics can partially explain why some spectral lines are not seen if thegas emitting the light is excited to temperature T . The probability that an given atom inthe gas will have its electron in energy level En is
℘(En) = D(En) e−EnkT
in which D(E) is the density of states; the number of configurations of energy E. This isalso called the degeneracy of the energy level E. Notice that this tells us that if En >> kTfor a given temperature, we are very unlikely to find any atoms in energy state En. If noatoms are to be found in that state, we will see no light emissions from atoms falling out ofthis state, and so many spectral lines emitted when atoms drop out of high energy stateswill not be seen unless the gas is hot enough to have an appreciable number of atoms in therequisite high energy starting-states to begin with.
Example At fairly low temperatures T , gas with the energy level spectrum of the previousexample may not be excited enough to have many atoms in E4, so we will see almost nolight of the frequencies f4,3, f4,2, f4,1 emitted. At low T this gas will have emission spectrumlike that shown below.
B RLow T Emission Spectrum
However we will see in our excursions into wave mechanics, a far more correct picture of theatom, that certain transitions between energy levels are forbidden on the basis of angular
96
momentum conservation, and other selection rules. Light carries angular momentum of1 h, called spin angular momentum, and so in the emission of light by excited atoms, cer-tain transitions might not be possible because the initial and final states may have angularmomenta that differ by the wrong amount for angular momentum conservation. This factwas not known at the time of Bohr-Sommerfeld theory, and cannot be properly taken intoaccount except within the framework of wave mechanics, the successor-theory to the Bohr-Sommerfeld model.
The inability of Bohr-Sommerfeld theory to explain the relative intensity of spectral lines wasone of its major failings. Different spectral lines are of different intensities because of severalfactors, one of which is the degeneracies of the two quantum energy levels involved in thetransition. Suppose that We have four quantum energy levels E1 < E2 < E3 < E4 and theyare degenerate. An electron in E1 can have angular momentum 1 h, and electron in E2 canhave angular momentum 1 h or 2 h and so one. We say that En is (n− 1)−fold degenerate.Let us couple this together with the notion that since lignt carries angular momentum 1 h,the only allowd atomic transitions must obey the selection rule
∆` = ±1, 0
We now list some transitions and the frequency of light emitted.
Initial N Initial ` Final N Final ` Allow/Forbid f4 4 1 1 forbid f4,1
4 3 1 1 forbid f4,1
4 2 1 1 allow f4,1
4 1 1 1 allow f4,1
4 4 3 3 allow f4,3
4 4 3 2 forbid f4,3
4 4 3 1 forbid f4,3
4 3 3 3 allow f4,3
4 3 3 2 allow f4,3
4 3 3 1 forbid f4,3
4 2 3 3 allow f4,3
4 3 3 2 allow f4,3
4 3 3 1 forbid f4,3
4 2 3 3 allow f4,3
4 2 3 2 allow f4,3
4 2 3 1 allow f4,3
4 1 3 3 forbid f4,3
4 1 3 2 allow f4,3
4 1 3 1 allow f4,3
.. .. .. .. .. ..
One can see the basic picture; there are two allowed transitions that will cause light offrequency f4,1 to be emitted, if the temperature is high enough, and ten transitions of emission
97
frequency f4,3 and so the ratio of intensities should be
If4,3 : If4,1 =10
2= 5
and we should see the spectral line of frequency f4,3 five times as bright at the f4,1 line.Bohr-Somerfeld theory does not give us the correct energy level degeneracies, and so itspredictions of spectral line intensities are not correct. However it does successfully predictthe actual frequencies of the spectral lines to excellent precision, and so many of its coreideas and assumptions must have some truth to them. The theoretical successor, wave me-chanics, is incomparably more accurate, more general, and more applicable to a wider rangeof problems.
10.3 Apparatus
1. Sodium light source. Gaertner Scientific.2. Hydrogen light source. Gas discharge tube powered up by a3. Variac. Type W5MT3.4. Diffraction grating.5. Spectrometer. Gaertner Scientific.
The figure shows how the setup will look, with the variac connected to the hydrogen dischargetube. The diffraction grating used in this lab (by me) has
m = 1, N = 25, 000, R = 25, 000
in whichmλ = d sin θ
R =∆λ
λ= mN
where R is the resolution in first order (m = 1). The quantity d in the first expression isthe distance between slits, the reciprocal of the number of lines per centimeter in the grating.
98
The first step is to determine the number of lines per centimeter by calibration using thetwo Sodium lines
λ1 = 5889.953A, λ2 = 5895.923A
We will first measure the angles at which the lines will be scattered by the grating.
θH source Slit colimator
Grating
Protractor (angle scale)
Telescopewith crosshair
Next we determine the angles at which maxima occur when light from hydrogen dischargetubes is passed through the grating. We then solve for the wavelengths by
mλ = d sin θ
Do not confuse m with an atomic quantum number, it is the diffraction order ofthe resolved spectral line.Finally we use the fact that for the Bohr model of hydrogen, emissions occur with wavelengths
E =hc
λ= −µc
2α2
2(
1
n2− 1
m2)
in which µ is the electron reduced mass µ = memp
me+mp≈ me and α = k
hcis the fine structure
constant
α = 0.0072974 ≈ 1
137
with k = e2
4πε0. Remember mec
2 = 0.511MeV and hc = 1239.8eV nm.There will only be a few visible wavelengths from which we will deduce
1
λ=
2π2µk2e4
h3c(
1
n2i
− 1
n2f
)
for ni → nf principle quantum number transitions, with
= R∞(1
n2i
− 1
n2f
)
99
the Rydberg constant R∞.What you should expect to see as you move the collimator from angle to angle is somethinglike the following
m=0 m=1m=-1 m=2m=-2
0 θ
10.4 Data and analysis
λ Trial 1 Trial 2 Trial 3
1 36.1833o 36.2o 36.1833o
2 36.233o 36.233o 36.232o
from which we obtain the averages
θ1 = 36.1833o, θ2 = 36.233o
Of course when you actually perform the experiment, you will need to do standard erroranalysis. These give line spacings for the grating of
a1 =5889.953A
sin(36.183o)= 9975.59A
and
a2 =5895.923A
sin(36.233o)= 9974.99A
for an average linespacing ofa = 9975.299± 558.76A
The Rydberg constant from Hydrogen spectra
With ni = 2 and nf = n;
color quantum number trial 1 (deg) trial 2 (deg) trial 3 (deg) average (deg)
red n = 3 40.1833 40.1833 40.2000 40.1889green n = 4 28.7167 28.750 28.7167 28.7278blue 1 n = 5 25.5500 25.5500 25.5667 25.5556blue 2 n = 6 24.5500 24.6333 24.6167 24.6000
100
From this data and a we find that
λ3 = 6437.16 A = 643.7nm, λ4 = 4793.82 A = 479.38nm, λ5 = 4303.24 A, λ6 = 4152.52 A
You can compare these with the exact, known values reported in the next experiment, agree-ment with known wavelengths is very good, to within 2%.These lines are transitions from the m = 2 level into one with quantum number in the table.
122 − 1
n2 n 1λ
.13889 3 .000155348
.18750 4 .000208601.2100 5 .000232382.22222 6 .000240817
The average value for the Rydberg constant from these four data can be gotten by plottingthe reciprocal wavelength versus the difference in the squared reciprocal quantum numberdifference, and doing a linear regression.
0.14 0.16 0.18 0.20 0.22 0.24 0.260.12
0.14
0.16
0.18
0.20
0.22
0.24
1000.0/λ
(n2 −
4)/4
n2
R∞ = 0.001057± 0.00010792A−1
which is in good agreement ( error is 4%) with the accepted value of
.00109737A−1
101
10.5 Diffraction gratings
Diffraction gratings are essentially multiple-slit aperatures that can be either reflection ortransmission devices. In a transmission grating, very fine grooves are cut into a transparentglass or polymer slide. These grooves act like aperatures and light transmitted through themundergoes multiple-source interference. From the figure below representing the grooves orcuts seen edge-on
θ
θ
r0
r1
r-1
dP
λ
we can use superposition to construct the intensity I at the point P on a screen upon whichthe transmitted light will fall. We represent a light wave of polarization P and wavelengthλ as
~E = pE0√r
sin (2π
λr − ωt) = p
1√rIm ei( 2π
λr−ωt)
and notice that if we label the central slit j = 0, then the j th slit is approximately
rj = r0 + jd sin θ
away from point P , where r0 is the distance from point P to the central slit, and d is theslit seperation. We can add up all of the waves arriving from all slits at point P ,
~ET =N∑
j=−N
pj
E0√rj
Im ei( 2πλ
rj−ωt)
≈N∑
j=−N
p0E0√r0
Im ei( 2πλ
rj−ωt)
102
≈ p0E0√r0
Im ei( 2πλ
r0−ωt)N∑
j=−N
ei 2πλ
jd sin θ
This sum is not difficult, callx = ei 2π
λd sin θ
then it isN∑
j=−N
xj = x−N (1 + x+ x2 + · · ·+ xN ) = x−N 1− xN+1
1− x
=xN+ 1
2 − x−(N+ 12)
x12 − x− 1
2
=sin ((2N + 1)πd sin θ
λ)
sin (πd sin θλ
)
and we can assemble the intensity from the time averaged Poynting vector
I(θ) =1
2µ0c
E20
r0(sin ((2N + 1)πd sin θ
λ)
sin (πd sin θλ
))2 = I0(
sin ((2N + 1)πd sin θλ
)
sin (πd sin θλ
))2
In this case our grating has 2N + 1 slits, the corresponding expression for an N -slit gratingis
I = I0(sin (Nπd sin θ
λ)
sin (πd sin θλ
))2
For N = 11 the intensity of transmitted light versus angle looks like the following figure.Notice that the bright fringes or maxima have an amplitude of about N 2 times the brightnessof the N−2 secondary maxima that lie between consecutive primary fringes. As N increasesthe disparity in relative brightness grows, until we see only very sharp, bright primary max-ima with complete darkness between them as N grows very large.
103
Diffraction grating, 11 slit
−2 −1 0 1 20
20
40
60
80
100
120
140
(πd sinθ)/λ
I(θ)
Typical gratings will be quoted as having a certain number of rulings or grooves cut intothem per centimeter. For example a grating with 10, 000 rulings per centimeter has a d-valueor slit-to-slit seperation of
d =1.0× 10−2m
104= 1.0× 10−6m = 1000.0nm
If we pass a light sample containing several distinct wavelengths of light through the grating,we will see one diffration pattern per wavelength present in the sample. These will not su-perimpose perfectly, and by measuring the spread we can determine the actual wavelengths.This is what is done in the Rydberg constant experiment.One must first establish the value of d for the grating being used by measuring the angularposition of a well known or standard wavelength of light, such as that from a laser or froma mercury lamp. Once d is known, the light to be measured is passed through the grating
104
and the angular positions of the various wavelengths present is measured.
Consider a spectrum containing two colors of visible light passed through a diffraction grat-ing, producing the intensity pattern seen below
Resolution in first order
−15 −10 −5 0 5 10 150
50
100
150
200
250
300
θ, degrees
I(θ)
We see two fringes resolved in first order since the m = 1 interference fringes for eachcolor are well separated. Carefully measuring with a ruler we see that the m = 1 fringeshave angular positions
λ θ (degrees)λ1 7.697λ2 9.236
105
Supose that we have previously measured our diffaction grating’s d value, and found that ithad 10, 000 rulings per centimeter, and so
d = 1000.0nm
Then since the positions of the primary maxima of the grating are
d sin θ = mλ
we discover thatλ1 = (1000.0nm) sin(7.697) = 133.9nm
andλ2 = (1000.0nm) sin(9.236) = 160.5nm
11 Measurement of unknown spectral lines
We can use the same apparatus to identify an unknown sample of gas from its visible emissionspectrum.This is done by creating a calibration curve for the instrument using a known sample, thebest being Mercury due to the number of visible spectral lines. The idea is that we use theinstrument to measure the angular location of particular spectral lines passed through thegrating, or through a prism, and match these lines to known wavelengths. We can make aplot of λ versus angle θ, but this will be nonlinear, so instead we will plot the sine of theangular displacement of a line versus λ.
Mercury spectral lines
Color λ, nm 1.0×106 nm−2
λ2
yellow-1 579.0 2.983yellow-2 577.0 3.004green 546.1 3.353blue-1 496.0 4.065 faintblue-2 491.6 4.138 faintblue-violet 435.8 5.265 faintblue-violet 434.7 5.265 faintblue-violet 433.9 5.265 faintdeep-violet 407.8 6.013 faintdeep-violet 404.7 6.106 faint
Hydrogen spectral linesred 656.3 2.322blue-green 486.1 4.232violet 434.0 5.309deep-violet 410.2 5.943
106
11.1 The Mercury data
Using the same apparatus as in the previous experiment, we calibrate the grating spectrom-eter with a Mercury source, using the four brightest lines.
182 50’
n=1, R n=2, Rn=1, Ln=2,L
θ displacement
I found the instrument to be centered at 182o50′ = 182.8333o, and the angular positions anddisplacements for the first three diffraction orders to be
Line Left Right Average Displacement θn = 1 first order of diffraction
violet 190.4 175.3 7.55green 192.333 173.333 9.5yellow-2 192.883 172.8 10.0417yellow-1 192.917 172.75 10.0833
n = 2 first order of diffractionviolet 198 167.55 15.225green 201.867 163.6 19.1333yellow-2 202.967 162.45 20.2583yellow-1 203.017 162.4 20.3083
n = 3 first order of diffractionviolet 205.667 159.6 23.0333green 211.667 153.217 29.225yellow-2 213.383 151.367 31.0083yellow-1 213.483 151.25 31.1167
In this table all angles have been converted to decimal degrees from degrees and minutes,for example 190o 24′ = 190.4o, 175o 18′ = 175.3o, and so forth.We now could compute the diffraction grating spacing a from
nλ = a sin θ
but we have already done such a thing in the previous experiment. If we were to do it nowfor this data and this grating, we obtain a = 33800 A.Instead we will correlate these lines with true wavelengths to produce calibration curvesrelating sin θ to λ for each diffraction order n, which we can use to determine the wavelengthof an unknown spectral line from its angular displacement.
107
11.2 First order calibration curve
Line λtrue nm Average Displacement θ rad sin θviolet 435.8 0.13177 0.13139green 546.1 0.16581 0.16505yellow-1 577.0 0.17526 0.17436yellow-2 579.0 0.17599 0.17508
A linear regression using the program lg in the support software or from the 499 websitearchive on rustam.uwp.edu on this λ versus sin θ results in a straight line
λ = 4.655966 + 3281.222900 sin θ
with a χ2 = 0.000151 for the fit, which is a very good fit.
11.3 Measuring Sodium lines
The Sodium source produces the following lines in the first order set of diffraction maxima;red; right angular position; 172o 46′, left; 194o 14′ with the instrument centered on 183o 30′.This results in an average angular displacement of θ = 10.7333o = 0.18733 rad, and wemeasure this lines wavelength from the calibration curve to be
λ = 4.655966 + 3281.222900 sin(0.18733) = 615.74nm
which is in superb agreement with the known red-Sodium line at λtrue = 615.43nm!The yellow Sodium line is seen in first order n = 1 diffraction at angular positions right;173o 14′ and right; 193o 47′, with an average displacement of θ = 10.275o = 0.179333 rad.We determine its wavelength to be
λ = 4.655966 + 3281.222900 sin(0.179333) = 589.94nm
again in very good agreement with the true value λtrue = 589.00nm.
108
12 Stefan-Boltzmann Radiation Law
The purpose is to verify the radiated power versus temperature for a blackbody radiator,
℘ = A∫ ∞
0
2πhc2
λ5(ehc
λkT − 1)dλ = σAT 4
in which A is the surface area of the radiator, σ is the Stefan-Boltzmann constant
σ = 5.67× 10−8 W
m2K4
12.1 Background
If a metallic box enclosing a cavity is slowly heated ,and the wavelengths of light emittedthrough a small hole in one wall is monitored, the proportion emitted at various wavelengthscould not be successfully predicted by classical physics. This is our first basic problem;computation of the fraction of energy radiated within a given wavelength range. The boxitself, unheated, is a perfect absorber since once light enters the hole its odds of bouncingback out are slim. The box is a perfect absorber since if it’s temperature is stable, any lightthat enters will be absorbed and re-emitted by the walls. A perfect absorber is also a perfectemitter (Kirchoff’s law), since light that enters the hole will be absorbed and re-emitted,and ultimately this energy must escape or the box temperature would rise.
Electromagnetic portion. The whole problem of heat radiation is complicated by thefact that it lies at the intersection of several very distinct fields of physics, in particularelectromagnetism and thermodynamics. We begin this investigation with its electromagneticcomponent.You are familiar with Maxwell’s equations in integral form from elementary physics
∮
SE · n dS =
qinS
ε0,
∮
SB · n dS = 0
and
d
dt
∫
SB · n dS = −
∮
C=∂SE · d`, µ0IthruC + µ0
d
dt
∫
Sε0E · n dS =
∮
C=∂SB · d`
In empty space both the static charge q and conduction current I, the currents of movingcharges, are zero, leaving Faraday’s and Ampere’s laws in a highly symmetrical form
d
dt
∫
SB · n dS = −
∮
C=∂SE · d`, µ0
d
dt
∫
Sε0E · n dS =
∮
C=∂SB · d`
We now apply the well known vector calculus identity
∮
C=∂SV · d` =
∫
∇×V · n dS
109
in which S is a two-dimensional surface in space, with normal n and right-handed boundingcurve C. This allows us to write both sides of the two dynamical laws of electromagnetismas
d
dtB = −∇×E, µ0ε0
d
dtE = ∇×B
To proceed further we apply another vector calculus identity∮
S=∂VV · n dS =
∫
V∇ ·V d3x
in which S is a closed surface enclosing volume V . We apply this to the two static equations(Gauss’s laws) to obtain, in empty space
∇ ·E = 0, ∇ ·B = 0
Maxwell used these four equations to prove that light is an electromagnetic wave, a factconfirmed experimentally by Hertz. To do this, first differentiate the two dynamical equationsto get
d
dt∇×B = −∇× (∇×E), µ0ε0
d2
dt2E =
d
dt∇×B
and substitute one into another to arrive at
−∇× (∇×E) = µ0ε0d2
dt2E
and now use a third vector identity to simplifiy the cross product
−∇× (∇×E) = −(∇(∇ ·E)−∇2E) = µ0ε0d2
dt2E
and use the differential form of Gauss’s law to arrive at the electromagnetic wave equa-tion
(∇2 − 1
c2∂2
∂t2)E = 0
in which the speed of propagation of the waves so constructed is
c =1√ε0µ0
= 3× 108 m
s
the previously established speed of light.We will now take for granted that light is an electromagnetic wave, and contemplate a hollowmetallic box containing electromagnetic standing waves, in thermal equilibrium with the hotwalls. By the time of Planck’s heat radiation experiments, it was widely accepted thatradiant heat was a form of electromagnetic radiation with long wavelenghts compared tovisible light.We now have two tasks; first to determine the spectrum of frequencies of standing ling wavesthat can exist inside the box, without being annihilated by interference, and second, todetermine how many distinct waves have the same frequency, or how many distinct waveshave frequencies between f and f + df . We will need to do this in order to compute the
110
spectrum of light emissions from a small hole cut in the wall.If the walls of the box are for example perfect conductors, so that no electric field canpenetrate into them, and if the walls are at x = 0, x = a, y = 0, y = a, and z = 0, z = a,then we must have E = 0 on the walls we find the solution
Ex = E0 sin(nπx
a) sin(
mπy
a) sin(
pπz
a)eiωt
for a cubic box of side a such that
(n2 +m2 + p2)π2
a2=ω2
c2
and each mode in the box is specified by these three integers n,m, p. For example in twodimensions let N(ω) equal the number of modes in a radius of r =
√n2 +m2 around zero
frequency. Then as r becomes very large the number of combinations(n,m) becomes thearea of a quarter circle.
−10 −5 0 5 10−10
−5
0
5
10
In two dimensions a light wave propagating an a given direction can have only one possiblepolarization direction perpendicular to its flow of energy, so
N(ω) = 1 · 14πr2 =
π
c2a2f 2
The number of modes per unit frequency interval is then
dN(f)
df= 4πf
a2
c2
In three dimensions we find that N is twice ( because any light wave propagating in direc-tion n can have two independent polarizations p such that p · n = 0 in three dimensions)
111
one-eigth the volume of a sphere of radius r and so
N(f) = 21
8
4
3πr3 =
8πa3 f 3
3c3
so that
dN
df=
8πa3 f 2
c3
This is referred to as the density of states for electromagnetic waves in the box.The next stage is to compute the total energy possessed by the radiation within the boxaccording to
dU
dV=U
a3=∑
f
dN
df· u(f) =
∫
d2U(f, T )
dV dfdf
in other words we multiply the number of modes of frequency f by the average energy ofsuch a mode when in equilibrium with the hot walls at T , and sum over all of the modes.This requires input from thermodynamics to compute the average energy that a light wave,which we will regard as an oscillator since it satisfies an oscillator equation, canextract from the walls at equilibrium.Thermodynamic portion. According to the classical equipartition of energy each quadraticdegree of freedom in the Hamiltonian has average energy of kT
2when in equilibrium with
a heat bath. Using the Gibbs Canonical Ensemble,
E =∫ ∞
0℘(E)E dE =
∫∞0
2πωE e−
EkT dE
∫∞0
2πωe−
EkT dE
= kT
This amount of energy then determines the amplitude of the wave, since in classical elec-trodynamics the energy density of electromagnetic radiation is proportional to the electricfield amplitude squared. The average energy stored as electromagnetic waves in the box pervolume per frequency interval is
d2U(f)
dfdV=dN(f)
dVu(f) =
8πkTf 2
c3
This is the Rayleigh-Jeans law. It is of course not in agreement with experiment, asPlanck was to show. This law predicts an ultraviolet catastrophy; the box would containan infinite total energy, since it contains ever increasing amounts of energy at higher andhigher frequencies.
112
Rayleigh−Jeans curves for T=T0, 1.5 T0, 2.0 T0
0 5 10 15 200
200
400
600
800
hf/kT
d2 U(f
,T)/
df d
V
The problem lies in the assumption that the equipartition law is valid. This cannot betrue, there must be dependence upon the frequency, or the theory is doomed to ultravioletcatastrophe.Planck noticed that the experimental data can be fit with tremendous accuracy by insteadassuming that each mode, which is in fact an oscillator by virtue of being waves,can only havea discrete spectrum of excitation energies rather than all possible energies, with anaverage determined by the temperature. He proposed that the energy of each mode is n (hf)in which n is 0, 1, 2... (the Planck hypothesis) and the constant h needs to be determined byfitting the experimental data.This assumption of discrete oscillator energies as opposed to continuous simply translatesinto the need to sum rather than integrate in order to compute the average energy per mode.We find now that
E =∫ ∞
0℘(E)E dE =
∫∞0
2πωE e−
EkT dE
∫∞0
2πωe−
EkT dE
→∑∞
n=0En e−En
kT
∑∞n=0 e
−EnkT
Fortunately this is no harder to compute than the integrals of the Equipartition version,since the sums are geometric
E =hf
∑∞n=0 n e
−nhf
kT
∑∞n=0 e
−nhfkT
=
hf e−
hfkT
(1−e−
hfkT )2
1
1−e−
hfkT
=hf
ehfkT − 1
113
This is the celebrated Bose-Einstein distribution function, and inserting this into theexpression for the energy density per frequency interval in the box yields
d2U(f)
dfdV=dN(f)
dVu(f) =
8πf 2
c3hf
ehfkT − 1
which is in exact agreement with the experimental data, and correctly predicts the totalenergy stored in the box as a function of temperature,the Stefan-Boltzman law
dU
dV=∫ ∞
0
d2U
dfdVdf =
8πhk4T 4
c3h4
∫ ∞
0
x3dx
ex − 1=
8πk4
c3h3
π4
15T 4 =
4
cσT 4
in which σ = 5.67× 10−8 Wm2 K4 is the Stefan-Boltzmann constant. The graph of d2U
dfdVversus
f has a maximum value at a wavelength predicted by the Wein law of displacement
Planck curves for T=T0, 1.5 T0, 2.0 T0
0 5 10 15 200
2
4
6
8
10
12
hf/kT
d2 U(f
,T)/
df d
V
λmaxT = 0.2012hc
kWhich can be found by maximizing the energy per frequency interval with respect to fre-quency.
The Stephan-Boltzmann law has been used to estimate the size of the universe as it cools,since cooling must be at constant total energy, assuming that the dominant constituent ofthe universe is the electromagnetic field filling it. From U = 4
cσ V T 4 we obtain the expansion
ruleV1 T
41 = V2 T
42
114
12.2 Luminosity or radiance
Consider now the problem of the luminosity of a small hole cut into the wall of the cavity.As a function of the temperature of the box, what is the total flow of power in the formof heat radiation through the hole? In the figure below we see a hole of area A. Considera volume element dV = r2dr sin θ dθ dφ within the cavity containing electromagnetic fieldenergy. The energy per unit volume in the cavity is
d2U
df, dV=
8πhf 3
c3(ehfkT − 1)
(Planck),8πf 2kT
c3(Rayleigh-Jeans)
and so the little volume element contains energy within freqency interval [f, f + df ] of
dU
df=
8πhf 3
c3(ehfkT − 1)
r2dr sin θ dθ dφ (Planck),8πf 2kT
c3r2dr sin θ dθ dφ (Rayleigh-Jeans)
all in the form of photons, all moving isotropically away from the box at speed c. The boxis a distance r < c dt away from the hole of area A.
n-z
dV
r
dr
θ
c dt=R
θ
The fraction of the radiant energy in the volume element heading in a direction n that willtake it through the hole is
angle subtended by hole
4π=
Az·nr2
4π=A cos θ
4πr2
and so the energy from this particular volume element that will leave through the holewithin time dt is
d∆Uthru
df=
8πhf 3
c3(ehfkT − 1)
r2dr sin θ dθ dφA cos θ
4πr2(Planck),
8πf 2kT
c3r2dr sin θ dθ dφ
A cos θ
4πr2(Rayleigh-Jeans)
115
We now add up the contributions from all of the volume elements within a hemisphere ofradius R = c dt against the wall of the box containing the hole; for the Planck case
d∆Uthru,total
df=
8πhf 3
c3(ehfkT − 1)
∫ R
0
∫ π2
0
∫ 2π
0r2dr sin θ dθ dφ
A cos θ
4πr2=
8πhf 3
c3(ehfkT − 1)
c dtA
4
and for Rayleigh-Jeans
d∆Uthru,total
df=
8πf 2 kT
c3
∫ R
0
∫ π2
0
∫ 2π
0r2dr sin θ dθ dφ
A cos θ
4πr2=
8πf 2 kT
c3c dtA
4
Now divide by dt and use P = dUdt
to get the power at frequencies between f and f + dfradiated through the hole
dP
df=
8πhf 3
c3(ehf
kT − 1)
cA
4Planck
dP
df=
8πf 2 kT
c3cA
4Rayleigh-Jeans
The ultraviolet catastrophe is now pretty obvious for Rayleigh-Jeans; integration over allfrequencies results in an infinite power flow for the Rayleigh-Jeans law, but results in thewell-known and correct Stephan-Boltzmann law for the Planck version
Ptotal =∫ ∞
0
dP
dfdf =
∫ ∞
0
8πhf 3
c3(ehfkT − 1)
cA
4df =
8π(kT )4
c3h3
π4
15
cA
4= σ AT 4
for the power flowing from the hole. This is true for all Blackbody radiators, such as themetal filament used in this experiment.
12.3 Apparatus
1. Stefan-Boltzmann Lamp. Pasco Scientific model TD-8555.
2. Thermal radiation sensor (thermopile). Pasco model TD-8553.3. 3 Subtronics model 2000 multi-meters, to be used as voltmeters and an ammeter.4. Power supply, 0-20 V. Micronta dual tracking DC power supply.
116
The apparatus is set up as in the figure.
Model TD-8555Stefan-Boltzmann
Lamp
CAUTIONVoltmeter
0-20 V Power
Ammeter
Voltmeter
It is very important that the voltage applied to the lamp never exceed 13 V, and that thecurrent through the lamp never exceed 3 A.
The power received by the heat sensor which detects thermal radiation in the range
0.5µm ≤ λ ≤ 25.0µm
will be monitored by placing the lamp a fixed distance from the detector. The lamp ispowered up and the voltage and current V , and I through the filament are recorded alongwith the detector voltage. The resistance
R =V
I
117
of the filament is a well known function of it’s temperature,
T =R− Rref
αRref
+ Tref
at low temperatures, and at high temperatures scales in a very precise way with respect toa reference resistance.
The first step in determining the temperature function is to measure the resistance of thefilament at room temperature. To do this, carefully measure the room temperature, hopefullyaround 300K, and measure the resistance R300 of the filament with a multi-meter. Whenthe filament is hot, its resistance will be related to the temperature by finding the ratio
R(T )
R300
and looking up the corresponding temperature in Kelvin in the table below, provided by themanufacturer.
118
RR300
Temperature (K)
1.0 3001.43 4001.87 5002.34 6002.85 7003.36 8003.88 9004.41 10004.95 11005.48 12006.03 13006.58 14007.14 15007.71 16008.28 17008.86 18009.44 190010.03 200010.63 210011.24 220011.84 230012.46 240013.08 250013.72 260014.34 270014.99 280015.63 290016.29 300016.95 310017.62 320018.28 330018.97 340019.66 3500
We could also interpolate the graph below, made from this data.
119
0 5 10 15 200
500
1000
1500
2000
2500
3000
3500
R/R_300
Temp.
Instead we will use a Lagrange interpolation program that has this table coded into it. Ituses
T (R[n] + δr) = −(δr − 1)(δr − 2)(δr − 3)
6T [n] +
δr(δr − 2)(δr − 3)
2T [n+ 1]
−δr(δr − 1)(δr − 3)
2T [n+ 2] +
δr(δr − 1)(δr − 2)
2T [n+ 3]
and is given below
/* temperature interpolator for SB experiment */
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
double R[34],T[34],alpha,y;
float r,c1,c2,c3,c4;
int n,m;
main(int argc, char *argv[])
R[0]=1.0;T[0]=300.0;
R[1]=1.43;T[1]=400.0;
R[2]=1.87;T[2]=500.0;
R[3]=2.34;T[3]=600.0;
120
R[4]=2.85;T[4]=700.0;
R[5]=3.36;T[5]=800.0;
R[6]=3.88;T[6]=900.0;
R[7]=4.41;T[7]=1000.0;
R[8]=4.95;T[8]=1100.0;
R[9]=5.48;T[9]=1200.0;
R[10]=6.03;T[10]=1300.0;
R[11]=6.58;T[11]=1400.0;
R[12]=7.14;T[12]=1500.0;
R[13]=7.71;T[13]=1600.0;
R[14]=8.28;T[14]=1700.0;
R[15]=8.86;T[15]=1800.0;
R[16]=9.44;T[16]=1900.0;
R[17]=10.03;T[17]=2000.0;
R[18]=10.63;T[18]=2100.0;
R[19]=11.24;T[19]=2200.0;
R[20]=11.84;T[20]=2300.0;
R[21]=12.46;T[21]=2400.0;
R[22]=13.08;T[22]=2500.0;
R[23]=13.72;T[23]=2600.0;
R[24]=14.34;T[24]=2700.0;
R[25]=14.99;T[25]=2800.0;
R[26]=15.63;T[26]=2900.0;
R[27]=16.29;T[27]=3000.0;
R[28]=16.95;T[28]=3100.0;
R[29]=17.62;T[29]=3200.0;
R[30]=18.28;T[30]=3300.0;
R[31]=18.97;T[31]=3400.0;
R[32]=19.66;T[32]=3500.0;
/* find R in table closest to input r */
n=0;
r=atof(argv[1]);
for(m=0;m<31;m++)
if(fabs(r-R[m]) <= fabs(r-R[n]))
n=m;
/* now we have bracketting values */
alpha=r-R[n];
c1=-(alpha-1.0)*(alpha-2.0)*(alpha-3.0)/6.0;
c2=alpha*(alpha-2.0)*(alpha-3.0)/2.0;
c3=-alpha*(alpha-1.0)*(alpha-3.0)/2.0;
c4=alpha*(alpha-1.0)*(alpha-2.0)/6.0;
y=c1*T[n]+c2*T[n+1]+c3*T[n+2]+c4*T[n+3];
printf("R/R(300) = %f at T = %f\n",r,y);
121
We simply run it by passing it the RR300
that we want T for. You will find this program as adownload on the 499 website archive.
12.4 Procedure
First measure the background reading of the thermal sensor with no power applied to thelamp. This background must be subtracted from all of the subsequent detector readings.
With the lamp and detector a fixed distance apart, apply a voltage to the lamp, wait untilall meter readings stabilize, and record filament current, voltage and detector voltage, whichis proportional to the power it receives within the given wavelength band.
12.5 Data and analysis
In 1996 I took the following data for this experiment.
Fil. V (volts) Fil. I (amps) Sns V (mvolts)
0.289 0.636 0.2200.867 0.933 0.8302.57 1.36 4.954.26 1.72 11.365.08 1.88 15.286.29 2.09 21.786.99 2.21 25.997.56 2.24 31.10
Fil. V (volts) Fil. I (amps) Sns V (mvolts)
Dark 0.043 0.144 -0.005Background 0.0005 0.001 0.03
This data is corrected with the awk script
awk ’print $1-0.0005, " ", $2-0.001, " ", $3-0.03’ data>newdata
and the resistances found with
awk ’print $1, "&", $2, "&", $3, "&", $1/$2, "\\\\", "\hline"’ newdata
which results in the table
122
Fil. V (volt) Fil. I (amp) Sns. V (mvolt) R (ohm)
0.2885 0.635 0.19 0.4543310.8665 0.932 0.8 0.9297212.5695 1.359 4.92 1.890734.2595 1.719 11.33 2.477895.0795 1.879 15.25 2.70336.2895 2.089 21.75 3.010776.9895 2.209 25.96 3.16417.5595 2.239 31.07 3.37628
The resistance at room temperature is gotten by measuring the current and resistancethrough the filament at very low applied voltage, essentially the ”dark” filament line ofdata above,
R300 =0.0005V
0.001A= 0.5Ω
and we then process our data file ”newdata”, the corrected voltages and currents, with
awk ’ print $3, "&", $1/(0.5*$2), "\\\\", "\\hline"’newdata
which gives us the table
Sns. V (mvolt) R(T )R300
T (K)
0.8 1.85944 5004.92 3.78146 87511.33 4.95579 110015.25 5.4066 117521.75 6.02154 130025.96 6.3282 136031.07 6.75257 1430
We now compare our results to the theoretical predictions. The total power radiated withinthe wavelength band sensed by the IR detector is
℘ = A∫ 25.0µm
0.5µm
2πhc2
λ5(ehc
λkT − 1)dλ = σAT 4
which we compute for each temperature using the following C program
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int n; /* for looping */
double lambda, dlambda, constant1, constant2, integral,factor;
123
float T;
/* constant1 = 2 pi h c^2 */
/* constant2 = hc/k */
main(int argc, char *argv[])
if(argc < 2 || argc >2)
printf(‘‘ ./sb temp’’\n);
exit(0);
T=(double)(atof(argv[1]));
dlambda=0.0000001; /* dlambda=1.0E-7 */
constant2=0.014413;
constant1=3.749176e-16;
integral=0.0;
for(n=50; n<=2500;n++)
lambda=(float)n*dlambda;
factor=lambda*lambda;
factor=factor*factor;
factor=factor*lambda; /* lambda^5 */
factor=factor*(exp(constant2/(lambda*T))-1.0);
integral=integral+1.0/factor;
integral=integral*dlambda*constant1;
printf(‘‘%f\t%f\n’’, T, integral);
Our sensor does not recieve all of the radiated power, but a fixed fraction depending on it’ssurface area and the distance from the filament, however the sensor voltage is proportionalto the received power, and so the plot of lnVsens versus ln℘ should be a straight line of slopeone, since by hypothesis they are proportional to one another. The C program gives the data
Vsense (mV) T (K) ℘theory
0.8 500 3350.554.92 875 32695.811.33 1100 82104.515.25 1175 10700221.75 1300 16053025.96 1360 19236931.07 1430 235243
The log-log plot of the theoretical power versus sensor voltage is below
124
0.1 1 10 100103
104
105
106
logV_sens
logP
which has a slope of 0.933, confirming within an accuracy of 7% the Stefan-BoltzmannRadiation law.
125
13 Measurement of the Compton Edge
The purpose of this experiment is to test the validity of the Compton Scattering expressionfor γ rays scattered through large angles (180o).
13.1 Background
The concept that light can be treated like a particle in some applications is not off the wall,there is evidence that in collisions with matter light behaves like any other particle. In theCompton effect a stationary electron is bombarded with photons and the scattering anglesand energies satisfy standard relativistic versions of energy and momentum conservationlaws.
γ
γ
φ
θ
m
m
The relativistic dispersion relation
E =√
p2c2 +m2c4
suggests that for massless particles such as light the magnitude of the momentum vector isE/c. The De Broglie relation
p =E
c=hc
λ
allows us to write the conservation of energy-momentum equation
pbefore = pafter
(hf
c, 0, 0, hf)+(0, 0, 0, mc2) = (
hf ′
ccosφ,
hf ′
csin φ, 0, hf ′)+(p cos θ,−p sin θ, 0,
√
p2c2 +m2c4)
This gives us two momentum conservation
hf
c=hf ′
ccosφ+ p cos θ, 0 =
hf ′
csin φ− p sin θ
and one energy conservation law
hf +mc2 = hf ′ +√
p2c2 +m2c4
126
Take the last relation and square it
(hf +mc2 − hf ′)2 = p2c2 +m2c4 = (hf − hf ′)2 + 2mc2 (hf − hf ′) +m2c4
and square the first two relations and add them
(hf
c− hf ′
ccosφ)2 = p2 cos2 θ, (
hf ′
csin φ)2 = p2 sin2 θ
to get
p2 = (hf
c− hf ′
ccosφ)2 + (
hf ′
csinφ)2 = (
hf
c)2 + (
hf ′
c)2 − 2h2ff ′
c2cos φ
orp2c2 = (hf)2 + (hf ′)2 − 2h2ff ′ cosφ
We have two equations for p2c2, equate them
(hf)2 + (hf ′)2 − 2h2ff ′ = (hf − hf ′)2 + 2mc2 (hf − hf ′)
reduce them algebraically to
2mc2(hf − hf ′) = 2h2ff ′(1− cosφ)
ormc2
h(f − f ′
ff ′ ) = (1− cosφ),mc2
h(
1
f ′ −1
f) = (1− cosφ)
and use cf
= λ to arrive at the Compton scattering formula
∆λ = λ′ − λ =h
mc(1− cosφ)
in which φ is the angle that the light is scattered through.
This equation makes a prediction about what direction light will be scattered into when itis involved in a collision with static electrons, and therefore can be verified (or invalidated)by experimentation. From our conservation law for energy we see that
h(f − f ′) =√
p2c2 +m2c4 −mc2 = Ek
which is exactly the kinetic energy of the electron. But we also know that
mc2
h(1
f ′ −1
f) = (1− cosφ)
which we can solve for f ′;
1
f ′ =1
f+
h
mc(1− cosφ) =
mc2 + hf(1− cosφ)
fmc2
127
or
f ′ =fmc2
mc2 + hf(1− cosφ)=
f
1 + hfmc2
(1− cosφ)
The electron kinetic energy is then
Ek = hf − hf
1 + hf
mc2(1− cos φ)
= hfhf
mc2(1− cosφ)
1 + hfmc2
(1− cosφ)
which is clearly maximized by a direct back-scattering of the light, φ = π for which 1−cosφ =2, giving us
Ek,max = hf2 hf
mc2
1 + 2 hfmc2
= hf1
1 + mc2
2hf
This is the maximal kinetic energy that a stationary electron can obtain in a collision witha photon, and is called the Compton edge energy Ee. It is easy to set up the experiment,and measure the stopping voltage needed to keep post-collision electrons from arriving at acollection plate, where they will complete a circuit and create a current in the apparatus. Allexperiments indicate that this result, gotten by assuming that light behaves like a particlein its interactions with matter, is correct.
A common version of this experiment is the measurement of the Compton Edge at whichthe photon is backscattered at angle θ = π. From our work above we see that this results inphotons of energy
E ′γ = hf ′ =
hf
1 + 2hf
mc2
=Eγ
1 + 2Eγ
mc2
The actual experiment is carried out with a γ-ray source, producing primary radiationof energy Eγ, a scintillation detector, and a multichannel analyzer. Gamma raysentering the detector undergo Compton scattering from electrons in a crystal (NaI dopedwith Thallium), liberating electrons. The detector produces electronic pulses of intensityproportional to the electrical energy dissipated in the crystal. Using a series ofphotoelectric dynodes, the scintillator produces a signal (photopeak) due to the primaryradiation, by using it to photoelectrically eject electrons. Another signal, the ComptonPlateau, is due to primary radiation scattering electrons out of the conduction band of thecrystal. The detector produces pulses proportional to the energy carried by these electrons,which of course can acquire any energy from the primary gamma ray, right up to the maximalvalue at the Compton edge
0 ≤ Ek ≤ Ek,max = hf1
1 + mc2
2hf
These signals are sorted by the MCA, which records the number of signals recieved versusintensity or channel. The channel ch of the MCA output is linearly related to the energyof the event that caused the signal
ch = a+ b E
128
and so the device can be calibrated by recording the channels of two gamma ray sources ofknown energy,
ch1 = a + bEγ,1, ch2 = a + bEγ,2
from which
b =ch1 − ch2
Eγ,1 − Eγ,2, a =
ch2Eγ,1 − ch1Eγ,2
Eγ,1 − Eγ,2
A typical MCA output looks like the following;
MCA output
0 200 400 600 8000
500
1000
1500
2000
2500
Channel ch
Cou
nts
The ideal output would be a sharp spike at a channel corresponding to the primary gammaray Eγ, and a sharp-edged plateau extending from 0 to Ek,max, but these features are broad-ened by secondary scattering and other phenomena. The actual Compton edge occurs abouthalf-way up the plateau (at ch ≈ 330 on the figure).
13.2 Apparatus
1. 137Cs sample.2. 60Co sample.
129
3. Multichannel Analyser. Canberra Series 30.4. NaI Scintillation detector.5. Power supply. Hewlett-Packard 0-200 V, 0-6 mA, 61516 A DC power supply.6. Lead bricks and sheeting.
on
off
Power Intensity Expand ScanEnter Clear HV Gate Signal
off
on
roll off on off on
Roi
%
Dead Time
4 2
N MTL
count
cursor
PHAN x10
M
MCSNx10
M+1
23 4 5
6
vertical
1/1
1/2 2/2
com
memoryvert.
X2I/OACQ
col timeadd subpha
mcsmcsr in
outsel out
tty ext
I/O
5121024
2048
ADC gain
ADC offset0
256
512 1024
2048
LLD ULD
high low
adc zero threshholdinput lev.
AmpgainX1
Lead bricks Scintillationcounter
Powersupply
120VAC
120VAC
channel
13.3 Scintillation counters and photomultipliers
When a large collection of atoms in a gas or liquid are condensed to form a solid (crystalline),the energy levels spread out to form energy bands as shown below. Valence electrons willfill the lower energy valence bands, two electrons per band level (spin up, spin down). Con-duction electrons will occupy the upper energy conduction bands, and be very delocalizedin the crystal, in a metal this gives rise to electrical conductivity. In an insulator thereare no electrons in the upper conduction bands, and there is an appreciable gap in energybetween valence and conduction bands, so it is very hard to promote a valence electron intoa conduction band level. In semi-conductors this band gap is very much smaller, so electronscan be promoted into the conduction bands much more easily.
130
gaseous (large r)crystalline (small r)r
E
E2
E1
The electrons with which photons produced by the 137Cs and 60Co gamma-ray sources usedin this experiment will collide belong to the NaI crystal doped with Thallium in the scintil-lation counter. A gamma ray photon can collide with an electron in the valence band of theNaI crystal, giving it enough energy to be ejected into the band of conduction energy levelsor even higher.
Conductionband
Band gap(6 eV for NaI)
Valenceband
Incident
ejected electron leaves a hole
Thalliumlevels
Eγ
The ejected electron leaves a hole in the valence band . Neighboring atoms try to shareelectrons in attempts to fill in the hole,and this causes the hole to move through the crystal.If the hole encounters a Thallium atom impurity, it can ionize the Thallium atom and thepreviously ejected electron can fall into one of the excited states of the thallium atom. Inthe drop from the conduction band intothe excited thallium electronic level, a photon of a
131
few eV energy is emitted, that starts the entire scintillation process.The electron ejected from the valence band by the γ-ray is called the primary electron,and doping with Thallium provides intermediate energy levels in the forbidden energy re-gion between valence and conduction bands. The primary electron rejoins the valence bandby falling into these intermediate levels, performing an otherwise forbidden transition andemitting light. The light produced is reflected by alumina or MgO reflective layers onto aphotocathode that emits electrons by the photoelectric effect.
NaI
Photocathode
Dynodes
The phototube then increases the number of liberated electrons 106 fold by producing sec-ondary electron emissions with a series of dynodes. This electronic pulseis delivered to anamplifier. To a very good approximation the pulse created is directly proportional to theenergy dissipated in the scintillator, and so the number of pulses displayed versus amplitude(channel or energy) corresponds closely to the energy spectrum of the primary radiation.
The photoelectric effect is used in an important bit of laboratory apparatus called the pho-tomultiplier. Schematically, they look like the following figure. This device has a windowcoated with a photosensitive material that ejects an electron when a photon of sufficientenergy hits the window. This electron is accelerated by around 150V by a high voltage
132
electrode, which it strikes. This is the first stage of a multi-stage collection of dynodes.The accelerated electron proceeds to the next, higher voltage dynode, where it liberatesmore electrons. The accelerated electrodes cascade deeper into the nest of dynodes, liberat-ing more electrons. Eventually the single electron has become 106 electrons, a typical gainfactor for PM tubes operating at about 2000V . This many electrons produces a signal bigenough for electronics to process and record.
Gamma ray
Photo-cathode
Dynode
Distributor(high voltage to dynodes)
High voltage input
Signal output
The probability that the origin photon entering the tube actually ejects an electron fromthe photocathode is called the quantum efficiency. For light in the visible range, a goodPM tube has a quantum efficiency of around 0.25.The figure above shows the spectrum recorded by a NaI(Tl) crystal when exposed to 0.6616MeV γ-rays. Ideally we should get a continuous signal extending from zero energy to
133
Kmax = Ee, and a sharp peak at Eγ for a single gamma ray energy incident on the scin-tillator. In practice the Compton continuum is smeared out, as is the photopeak, due tosecondary collisions and e+e− pair creations. The Compton Edge or the channel at whichelectrons are ejected with Kmax = Ee by the γ-ray, is halfway up to the Compton plateau.This is the energy that we are looking for in this lab.
The Multichannel Analyser displays 1023 channels and plots the number of gamma raysdetected per channel as a graph. The relation between the channel at which the photon isdetected and the corresponding energy deposited by the event in the scintillator is linear
ch = a + bE
in which ch is the channel, E is the energy and a and b are constants to be determined bycalibrating the device as illustrated in a previous section.
For this we use the 60Co source which emits γ−rays of energies
E1 = 1173.210± 0.002KeV
andE2 = 1332.470± 0.002KeV
To calibrate the MCA, we will count the cobalt source for about 4 hours, and the countnumber versus channel will look like the figure below.
13.4 Stage 1; calibration
When the experiment was performed in 1995, the following was found; the energy E1 pho-topeak occured at
ch1 = 837, 11010 counts
andch2 = 947, 8408 counts
We compute the standard deviations of the values of ch1, ch2 by looking for the channels oneither side of the peak that hold half of these numbers of counts,
ch11 = 797, ch12 = 873
ch21 = 907, ch22 = 985
Assuming that each peak is a normal curve, the standard deviations given by
1
2A = Ae−
(ch−chp)2
2σ2
or, the half-height channels are deviated from the photopeak channels by
ch = chp ±√
2 ln 2σ = chp ± 2.35σ
134
and so
σch1 =ch12 − ch11
2.35= 32.34, ch1 = 837± 32.34
andch2 = 947± 33.19
We then set up the linear equations
837 = a+ b(1173.210), 947 = a+ b(1332.470)
and determine that the calibration constants are
a = 26.67, b = 0.6907
13.5 Stage II; photopeak and edge measurements
The 137Cs sample was then set up and the following photopeak spectrum was observed aftercounting for 30 minutes.
The main photopeak corresponding to energy
E3 = 661.64keV
was found withch3 = 476, 60, 000counts
ch31 = 453, ch32 = 496
and soch3 = 476± 18.3
and a Compton edge atche = 345± 10
corresponding to an energyEedge = 461keV
which is quite close to the accepted value of 477.318 keV . The photopeak corresponds to aγ-ray of energy
Eγ = 650.54 keV
agreeing well with the 661.64 keV accepted value.
14 Energy versus Momentum for Relativistic Electrons
When an electron in a scintillators valence band is struck by a γ-ray from a radioactivesource, the electron can acquire energies of several MeV, making them highly relativistic.By examining the energy of the Compton edge and the photopeak from the scintillator, onecan easily verify the form of the relativistic dispersion law
E2 = p2c2 +m2c4
135
14.1 Apparatus
The same as the Compton edge experiment, with additional 22Na, 137Cs, 54Mn, and 60Coγ-ray sources. The energies Eγ of the γ rays emitted by each source can be found below.
14.2 Procedure
The Compton edge and photopeak energies of all of the samples are measured. The actualphotopeaks should occur at
Isotope Eγ
22Na 0.511 MeV137Cs 0.622 MeV54Mn 0.835 MeV60Co 1.178 MeV
From this data the channel (ch) versus energy (E) calibration constants for the multichannelanalyser are determined
ch = a + bE
We found in the last expriment that the electron kinetic energy at the Compton edge is
Ee =2E2
γ
mc2 + 2Eγ
and at the Compton edge, the backscattered photon has energy
Eγ′ =mc2Eγ
mc2 + 2Eγ
Adding we find thatEe + Eγ′ = Eγ
equaling the energy of the pre-collision photon. Since the photon is backscattered we findthat
Pe =Eγ − (−Eγ′)
c
Eliminating Eγ′ we find that after the collision the electron momentum is
Pe =2Eγ − Ee
c
where Ee is the Compton edge energy and Eγ is the photopeak energy.The total energy of the electron is kinetic plus rest, or Ee +mc2, which presumably satisfies
(Ee +mc2)2 = P 2e c
2 +m2c4 = E2e + 2Eemc
2 +m2c4
136
or
E2e + 2mc2Ee = P 2
e c2,
P 2e
2m= Ee +
E2e
2mc2
In order to determine the true dispersion law, we will measure Ee and Eγ and plot P 2e
2meversus
Ee, assuming thatP 2
e
2me
= a0 + a1Ee + a2E2e
and run a nonlinear least squares fit to determine a0, a1, a2. Classically we should find thata0 = 0, a1 = 1, and a2 = 0, however what we will see is relativistic departures from thisdispersion law.
14.3 Data and analysis
The channels and corresponding energies for photopeaks and Compton edges for all foursources are in the table below.
Isotope cγ Eγ MeV ce Ee MeV22Na 107.5 0.511 69.5 0.341137Cs 143.0 0.622 93.7 0.44754Mn 183.5 0.835 137.7 0.64060Co 260.0 1.178 212.9 0.969
We are using the MCA calibration constants
E = (0.00438± 0.000035 MeV )c+ (0.0366 MeV )
found by matching the photopeaks to their known energies. We now compute the ”classicalenergy”
Isotope Ee MeV P 2e
2meMeV
22Na 0.341 0.453778137Cs 0.447 0.62153554Mn 0.640 1.0380660Co 0.969 1.88236
We now run a nonlinear least squares fit of this data to try to fit the best curve to it. I usedthe program polyfit to perform the regression, obtaining
P 2e
2me
= 0.002837 + 0.957349Ee + 1.015000E2e
in extremely good agreement with a supposition of a relativistic dispersion relation. Weobtain the estimate
mec2 =
1
2(1.015)MeV = 0.493MeV
which is in error by only 3.6%.
137
15 Michelson-Interferometer and the Index of Refrac-
tion of Air
Apparatus
1. Beck Interferometer M300/6407.2. Vacuum pump, air cell.3. He-Ne Laser.4. Phototransistor.5. Chart recorder.
The department has it’s own Michelson interferometer, however this apparatus belongs tothe chemistry department and is used in the physical chemistry lab course. The Beck inter-ferometer used in the Michelson mode looks like the following.
138
M2
M1
M3
Compensator
AirCell
To Vacuum pump
Source
Operation of the interferometer is simple. The eye is aligned with the viewing axis about 10inches from the instrument. There will be three reflected images of a pointer in the beamdivider head, as in A below.Adjust the tilt controls on M1 to see which image can be moved, and the control should beadjusted so that the moveable image superimposes on the stationary one B.
139
A B
C D
When exact vertical on horizontal coincidence of the images is achieved the interferencefringes will become visible. For unequal path lengths and nonparallel mirrors, the fringeswill be sections of circles surrounding a center displaced from the field of view. If the mirrorsare parallel but the path lengths are unequal, the fringes will be circular with a center withinthe field of view, order n in the middle, order n− 1, n− 2 and so forth of increasing radii. Ifthe path lengths are equal to within ±5 fringes, interference fringes will even be visible withwhite light from a tungsten filament.Zero path difference can be calibrated with a Mercury vapor lamp. Turn the path lengthcontrol dial in the upper right corner of the top figure in whatever direction causes thefringes to move inward, in the direction of decreasing radius. As the path difference becomeszero, the fringes contract to a central point. Now adjust the tilt controls to produce about10 nearly vertical fringes as in D, and adjust the path length control until the fringes arestraight. Now switch to a tungsten lamp and continue to turn the path length control veryslowly until a group of bright fringes with a distinctive central dark band in the middleappears in the field. Center this collection of fringes. The instrument is now ready and thepath length is zero.
Procedure
The He-Ne laser is split and recombined after passing through a pressurized cell. A photocelland chart recorder are used to count the number of wavelength shifts the insrutment cyclesthrough as the cell is evacuated, and then slowly repressurized.
Sample data
The length of the air cell used is 0.10m, the laser has a wavelength of
λ = 632.8173× 10−9m
140
We find that there will fit
2L
λ=
0.2m
632.8173× 10−9m= 316, 047 = N
waves in the cell in vacuum, and so
2L = Nλv =Nc
ν= ((N +N ′)λair) =
(N +N ′)c
nairν
where N ′ is the number of wavelengths the interferometer cycles through as the cell is refilled.Four separate trials revealed 83 fringes were shifted through as the cell was refilled with air,and so
nair =316, 047 + 83
316, 047= 1.0002626
which agrees very well with the accepted value of 1.0002684 at 70oF , the air temperature ofthe experiment.
141
16 Angular distribution of emitted radiation
The energies, and therefore frequencies, of light emitted by the Hydrogen atom when itundergoes a transition from one energy level to another, can be gotten from the Bohr formula
hν = ∆E = −Z2mc2α2
2(
1
N2i
− 1
N2f
)
however not all energy level transitions that one might naively assume occur do not in factever happen, because the transition violates certain conservation laws. We not only have anenergy balance to preserve
Ei = Ef + hν
but angular momentum must be preserved as well as linear momentum. The problem ofenergy emission by the atom is most appropriately addressed through quantum field theory,a theory developed to study processes in which particle number or number of quanta is notpreserved. Emission of light is a creation process,
|n, `,m〉 → |n′, `′, m′〉 ⊗ |k, σ〉
in which the rightmost ket is that describing a photon of wavevector k and polarization stateσ.We can treat this problem using perturbation theory and the Golden Rule developed earlier,if we model the interaction between the photon and the atom with
HI =∫
A · J d3x
in which J = −q h2im
(Ψ∗∇ − (∇Ψ∗)Ψ) is a “field operator” for the electronic current andA is the vector potential of the photon field. Both of these field operatorscan be Fourierexpanded in terms of solutions to the appropriate Schroedinger or Maxwell equation, and wewill do so in a very schematic manner, because we are in fact getting far ahead of ourselvesright now as far as subject matter is concerned.Let a†n,`,m create an electron with wavefunction ψn,`,m(r, θ, φ) from the vacuum, and an,`,m
be the corresponding annihilation operator, so that
[an,`,m, a†n′,`′,m′] = δn,n′δ`,`′δm,m′
and let b†k,σ create a photon with wavevector k and (transverse) polarization vector vk,σ.Recall that in empty space the independent polarization vectors must be perpendicular tothe wavevector
vk,1 · k = 0, vk,2 · k = 0, vk,σ · vk,σ′ = δσ,σ′
The part of a properly quantized photon vector potential responsible for creating a plane-wave photon will then look like
A =∫
d3k′∑
σ′=1,2
Nk′,σ′vk′,σ′ eik′·r b†k′,σ′
142
which when acting on the vacuum creates a photon with wavefunction
〈k, σ|A|0〉 =∫
d3k′∑
σ′=1,2
Nk′,σ′vk′,σ′ eik′·r 〈k, σ|k′, σ′〉 = vk,σ eik·r
There will be a similar Fourier expansion for the electron field
Ψ =∑
n′,`′,m′
(ψ∗n′,`′,m′(r, θ, φ)an′,`′,m′ + ψn′,`′,m′(r, θ, φ)a†n′,`′,m′)
When acting on the vacuum, this will create a Hydrogenic bound state particle with wave-function
〈n, `,m|Ψ|0〉 =∑
n′,`′,m′
ψn′,`′,m′(r, θ, φ)〈n, `,m|n′, `′, m′〉 = ψn,`,m(r, θ, φ)
We have performed a rather subtle conversion to the Bargmann-Fock space in which statesare labeled by the number of quanta of each type occupying the universe, and wavefunctionsare matrix elements of field operators in the basis of such particle number states.
We can now apply Fermi’s Golden Rule which states that the probability of transition frominitial state |i〉 to final state |f〉 is proportional to the modulus of the matrix element 〈f |HI|i〉.We find then that allowed transitions resulting in the emission of light will have nonzerovalues of ∫
d3x 〈nf , `f , mf ; k, σ|A · J|ni, `i, mi〉
When we exand J in terms of annihilation/creation operators, we only need to reatin theone term in four that has a nonvanishing matrix element between the one electron in andout states, we find then that the expression above reduces to
∫
d3x∑
n′,`′,m′
∑
n′′,`′′,m′′
∫
d3k′∑
σ′=1,2
Nk′,σ′e−ik′·r (− qh
2mi)vk′,σ′ ·(ψ∗
n′,`′,m′∇ψn′′,ell′′,m′′−∇ψ∗n′,`′,m′ψn′′,`′′,m′′)
·〈nf , `f , mf ; k, σ| bk′,σ′a†n′,`′,m′an′′,`′′,m′′ |ni, `i, mi〉which we now evaluate using
〈nf , `f , mf ; k, σ| bk′,σ′a†n′,`′,m′ = 〈0|δnf ,n′δ`f ,`′δmf ,m′δk,k′δσ,σ′
andan′′,`′′,m′′ |ni, `i, mi〉 = |0〉δni,n′′δ`i,`′′δmi,m′′
and we finally obtain the matrix element
℘ ∝ |∫
d3x e−ik·r(− qh
2mi)vk,σ · (ψ∗
nf ,`f ,mf∇ψni,`i,mi
−∇ψ∗nf ,`f ,mf
ψni,`i,mi)|2
which is basically just the Fourier transform of the electronic current matrix element betweenthe initial and final electron states.
143
The Schroedinger equation itself allows us to evaluate this matrix element in the dipoleapproximation,
e−ik·r ≈ 1
within the volume over which the bound state wavefunction probability density is apprecia-ble. For two different solutions to the Schroedinger equation
(− h2
2m∇2 + V (r))ψ∗
f = Efψ∗f , (− h2
2m∇2 + V (r))ψi = Eiψi
we multiply the first by ψi and the second by ψ∗f and subtract to obtain
− h2
2m∇ · (ψi∇ψ∗
f − ψ∗f∇ψi) = (Ef − Ei)ψ
∗fψi
Multiply both sides and integrate
− h2
2m
∫
d3x r∇ · (ψi∇ψ∗f − ψ∗
f∇ψi) = (Ef − Ei)∫
d3x rψ∗fψi
However for any vector f that vanishes sufficiently strongly at infinity;
∫
r∇ · f d3x =∫
(x, y, z) (∂fx
∂x+∂fy
∂y+∂fz
∂z) d3x
which can be integrated by parts to give
= (xfx|∞−∞, yfy|∞−∞, zfz|∞−∞)−∫
(fx, fy, fz) d3x = −
∫
f d3x
Applying this now to our current operator shows that
∫
(ψi∇ψ∗f − ψ∗
f∇ψi) d3x =
2m(Ef − Ei)
h2
∫
ψ∗f rψi d
3x
the transition probability is then, in the dipole approximation
℘ ∝ |2m(Ef − Ei)
h2 (− qh
2mi)vk,σ ·
∫
d3x (ψ∗nf ,`f ,mf
rψni,`i,mi) |2
which is proportional to the matrix element of the dipole operator qr betweenthe initial and final states.
Transitions resulting in the emission of light between two states ψi and ψf are said to beforbidden if the matrix element of the electric dipole operator between these two states iszero. Whether or not this matrix element is zero has little or nothing to do with the radialpart of the wavefunction, it is the angular wavefunctions that really determine the selectionrules since
x = r cosφ sin θ = −r√
8π
3<e Y1,1(θ, φ)
144
y = r sin θ sinφ = −r√
8π
3=mY1,1(θ, φ)
and
z = r cos θ = r
√
4π
3Y1,0(θ, φ)
and we can use the Clebsch-Gordon recursion relations deried much earlier in our study ofangular momentum to deduce that
x|ni, `i, mi〉 = A|nα, `i + 1, mi ± 1〉+B|nβ, `i − 1, mi ± 1〉
y|ni, `i, mi〉 = A′|nα, `i + 1, mi ± 1〉+B′|nβ, `i − 1, mi ± 1〉and
z|ni, `i, mi〉 = A′′|nα, `i + 1, mi〉+B′′|nβ, `i − 1, mi〉These tell us that the final state of the atom must differ from the initial by
`f = `i ± 1
andmf = mi ± 1, mf = mi
These are known as the electric dipole selection rules and they govern what we see inthe spectrum of atomic transitions. You may have noticed that Fermi’s Golden Rule evengives us some indication as to the polarization state of the light emitted in a given transition.
16.1 Polarization and angular distributions
Using the formula
℘ ∝ |2m(Ef − Ei)
h2 (− qh
2mi)vk,σ ·
∫
d3x (ψ∗nf ,`f ,mf
rψni,`i,mi) |2
we can determine the most probable direction of emission for an oriented atom or nucleusundergoing a transition, and we can determine the probability that a photon emitted in acertain direction n = (sin θ cosφ, sin θ sin φ, cos θ) will have a given polarization state. Forlight with k-vector k = kn, there will be two independent polarizations vk,σ, σ = 1, 2, whichwe will take to be
vk,1 = (cos θ cos φ, cos θ sinφ,− sin θ)
andvk,2 = (− sin φ, cosφ, 0)
which satisfy vk,σ · k = 0.If our atom or nucleus begins in a state with wavefunction
ψi(r, θ, φ) = Ri(r)Y0,0
145
we have seen from the dipole selection rules that the final state after photon emission mustbe
ψf (r, θ, φ) = Rf(r)Y1,m
with m = 0,±1. If we use
x = r
√
2π
3(Y1,1 + Y1,−1), y = −ir
√
2π
3(Y1,1 − Y1,−1), z = r
√
4π
3Y1,0
then our matrix elements will be∫
d3x (ψ∗nf ,`f ,mf
xψni,`i,mi) =
1√6
∫ ∞
0R∗
i (r) rRf(r)r2 dr(δm,1 + δm,−1)
∫
d3x (ψ∗nf ,`f ,mf
y ψni,`i,mi) =−i√
6
∫ ∞
0R∗
i (r) rRf(r)r2 dr(δm,1 − δm,−1)
and∫
d3x (ψ∗nf ,`f ,mf
z ψni,`i,mi) =
1√3
∫ ∞
0R∗
i (r) rRf(r)r2 dr δm,0
and we will use the abbreviation
Rif =∫ ∞
0R∗
i (r) rRf(r)r2 dr
We can now construct the probability that an oriented atom will emit light in the k = kndirection with polarization states 1 or 2. First we compute
vk,1·∫
d3x (ψ∗nf ,`f ,mf
rψni,`i,mi) = Rif ((cos θ cosφ)
1√6(δm,1+δm,−1)+(cos θ sinφ)(
−i√6)(δm,1−δm,−1)
− sin θ1√3δm,0)
=Rif√
6( cos θe−iφδm,1 + cos θeiφδm,−1 −
√2 sin θδm,0)
and for the other polarization
vk,2 ·∫
d3x (ψ∗nf ,`f ,mf
rψni,`i,mi) = Rif (− sinφ
1√6(δm,1 + δm,−1) + cosφ(
−i√6)(δm,1 − δm,−1)
=Rif√
6(− ie−iφδm,1 + ieiφδm,−1)
If we place a detector to receive light emitted in the n direction, regardless of polarization,the probability that the transition into the m = 0 state will result in light of any polarizationarriving at the detector is
∑
σ=1,2
℘0→0 = |2m(Ef − Ei)
h2 (− qh
2mi)|2 |Rif√
6|2 | −
√2 sin θ|2
146
and the probability that the transition into the m = −1 state will result in light of anypolarization arriving at the detector is
∑
σ=1,2
℘0→−1 = |2m(Ef − Ei)
h2 (− qh
2mi)|2 |Rif√
6|2 (| cos θeiφ|2 + |ieiφ|2)
= |2m(Ef − Ei)
h2 (− qh
2mi)|2 |Rif√
6|2 (
1 + cos2 θ
2)
and for the m = 1 we get the same result
∑
σ=1,2
℘0→+1 = |2m(Ef − Ei)
h2 (− qh
2mi)|2 |Rif√
6|2 (| cos θeiφ|2 + | − ieiφ|2)
= |2m(Ef − Ei)
h2 (− qh
2mi)|2 |Rif√
6|2 (
1 + cos2 θ
2)
If the transitions were from levels or into levels with ` 6= 0, 1, we would of course have dif-ferent matrix elements, and could end up with more complicated angular dependencies onour emission probabilities.
The rather good text Experiments in Modern Physics by Melissinos gives the following con-crete example, for which we can use our computations above. Consider a nucleus or atomthat decays from one quantum state to another in two stages as shown below.
(l,m)=(0,0)a
b
c
(l,m)=(1, m), m=0,-1,+1
(l,m)=(0,0)
ab transition
bc transition
The intermediate state has three possible orientations with respect to an arbitrary quan-tization axis, about which we have no knowledge. The first γ for a → b will be emittedisotropically.If we set up a detector as illustrated below, the direction between the sample of emitters andthe detector, receiving the first photon, establishes a quantization axis which we take to bethe z-axis. When this detector gets a γ of energy Ea − Eb, it came from a nucleus left withits angular momentum vector having z-component m = ±1 with respect to this axis (sincesin2 0 = 0). Now this nucleus performs its second decay, which must be a ∆m = −1→ 0 or∆m = 1→ 0, and so has an anisotropic distribution
℘(θ) ∝ (1 + cos2 θ
2)
or℘(θ)
℘(π2)
= (1 + cos2 θ
2)
147
16.2 Angular correlations of γ emissions
What we have said above applies to an oriented emitter. How can we orient an atom ornucleus? We could freeze it down to quench rotations and vibrations, and apply a strongmagnetic field. In the nuclear γγ angular correlation experiment, we will use nuclei thatemit two gamma rays. The direction of the first emitted γ will be used as the z-axisfor a study of the angular distribution of the second γ.Cobalt-60 decays to Nickel-60, which in turn emits first a 1.172MeV γ and then a 1.333MeVγ, with a delay of only 1× 10−12 s between these emissions.
Photomultiplier 1
Photomultiplier 2
θz
n
Gamma source
The decays that produce these γ rays are between high-angular momentum levels, and a de-tailed computation of the angular distribution of the second photon would yield a theoreticalangular distribution
℘(θ) = 1 +1
8cos2 θ +
1
24cos4 θ
This could be measured by moving the angular placement of a second photomultiplier tubefrom one angular position θ to another, 10− 15 degrees apart, and counting at each locationfor equal times.
How do we know that when one γ of the first emission energy arrives at the fixed counter(photomultiplier 1), and when the second arrives at the moveable counter, that they cmefrom the same nucleus? The signals from both PM tubes are fed into a coincidence circuit,which only registers simultaneous γ arrivals. The odds are pretty good that simultane-ously arriving γs of the right energy are correlated, meaning related; coming from the samesource. The delay for 60Co is about 1 × 10−12 s, and so the circuit must be tuned to countcoincident γ’s with this time delay between arrivals at the respective PM tubes.
148
16.3 A simulated experiment
I have not performed this experiment myself, but we can easily simulate it to see preciselywhat the data should look like.Apparatus would consist of1. Coincidence counter.2. Two photomultiplier tubes, equidistant from a 60Co source. Tube 1 is fixed, tube 2 is ona traveller that can be stopped at 10 degree position intervals.We will need to know the distance of the PM tubes from the source, in order to computethe angular width of the tubes as subtended from the source location. The PM tubes eachsubtend an angle dθ = 8.0 degrees as measured from the source.The coincidence circuit will count all events within a certain channel number of the photo-peak. This means that it is integrating the signal between one full peak standard deviation oneither side of the peak. The peak occurs at a channel corresponding to energy Eγ = 178 keV ,and we integrate then from 150 to 201 keV for each PM. The counting time is set at 100 sfor each angular position, and it gives us the following data, the angle θ is the position ofthe moveable PM.
Angle θ PM-1 integration PM-2 integration C(θ) coincident counts9.000000 13471 13555 1418.000000 13315 13433 1027.000000 13173 13208 936.000000 13105 13378 1145.000000 13254 13423 1454.000000 13420 13204 1663.000000 13286 13291 2072.000000 13146 13413 2681.000000 13465 13204 2690.000000 13249 13590 3699.000000 13454 13395 45108.000000 13451 13314 73117.000000 13417 13394 90126.000000 13217 13450 88135.000000 13422 13127 121144.000000 13269 13358 199153.000000 13260 13247 339162.000000 13086 13340 604171.000000 13221 13350 1386180.000000 13555 13570 2967
The simulation confirms the hypothesis that simultaneously emitted γ rays are emitted fromthe same nucleus, at at 180 degrees apart, consistent with conservation of momentum forpair annihilation.The program used for the simulation is listed below. It uses the accept-reject algorithm toproduce a random number sharply peaked (Lorentzian peak) about zero, which is used to
149
create a direction for the second photon once the first photon random direction has beencomputed.
/* gamma-gamma correlation simulation */
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define PI 3.1415926
#define dtheta 8.0 /* angular width in degrees */
long seed;
double d,dp,theta,del;
int n,m;
/* d, dp are two correlated directions */
/* dtheta is angular width of detectors */
/* theta is moveable detector position */
long c,c1,c2;
/* ci is counter i tally, c is coincidence tally */
double lorentz();
int trigger1(double angle); /* detect if PM is triggered */
int trigger2(double angle, double pos);
/* get a lorentzian random between 0, 360 peaked at 180 with respect to th*/
main()
seed=1342;
srand(seed);
del=9.0; /* angular increments for moveable PM */
/* counters set to 0 */
for(m=1;m<=20;m++)
theta=(double)m*del;
c=0;
c1=0;
c2=0;
for(n=0;n<300000;n++)
d=360.0*(double)rand()/(double)RAND_MAX;
dp=180.0+d+lorentz();
while(dp>360.0)dp=dp-360.0;
/*printf("%f\t%f\t%f\n", d,dp,fabs(d-dp));*/
if(trigger1(d)==1 && trigger2(dp,theta)==1)
c=c+1;
150
if(trigger1(dp)==1 && trigger2(d,theta)==1)
c=c+1;
if(trigger1(d)==1 || trigger1(dp)==1)
c1=c1+1;
if(trigger2(d,theta)==1 || trigger2(dp,theta)==1)
c2=c2+1;
printf("%f & %d & %d & %d\\\\\n", theta,c1,c2,c);
/*end loop over angles */
int trigger1(double angle)
if(angle >= 0.0 && angle <= dtheta)
return(1);
else
return(0);
int trigger2(double angle, double pos)
if(angle >= pos-dtheta/2.0 && angle <= pos+dtheta/2.0)
return(1);
else
return(0);
double lorentz()
double num,num2,test;
/* get a random angular change between -180, 180 */
/* peaked sharply around 0 */
start:
num=180.0*(2.0*(double)rand()/(double)RAND_MAX-1.0);
test=100.0/(100.0+num*num);
num2=(double)rand()/(double)RAND_MAX;
if(num2<=test)
return(num);
else
goto start;
double y11cor()
double num,num2,test;
/* get a random angular change between -180, 180 */
/* peaked sharply around 0 */
start:
151
num=180.0*(2.0*(double)rand()/(double)RAND_MAX-1.0);
test=0.5*(1.0+cos(num)*cos(num));
num2=(double)rand()/(double)RAND_MAX;
if(num2<=test)
return(num);
else
goto start;
A second unused routine is provided to generate random angles in degrees with PDF f(θ) =12(1 + cos2 θ).
We now work up our data by producing a polar plot of coincidence versus angle, whichtranslates into coincidence being the radial coordinate. The therefore plot the points
(x, y) = (℘(θ) cos(θ), ℘(θ) sin θ) ∝ (C(θ) cos(θ), C(θ) sin θ)
Which should be a spike directed to the left from the origin.
22Na γ γ correlations
−1.0 −0.8 −0.6 −0.4 −0.2 0.00.0
0.2
0.4
0.6
0.8
1.0
C(θ) cos(θ)
C( θ
) si
n(θ)
As a good exercise you are urged to modify the simulation to produce data consistent withthe angular correlations of emissions from the Cobalt source. The polar plot should look likethe following.
152
60Co γ γ correlations
−1.0 −0.5 0.0 0.5 1.00.0
0.5
1.0
1.5
2.0
C(θ) cos(θ)
C( θ
) si
n(θ)
and those for the example used in the computations above, with
C(θ) =1
2(1 + cos2 θ)
would look like the following;
dipolar γ γ correlations
−1.0 −0.5 0.0 0.5 1.00.0
0.5
1.0
1.5
2.0
C(θ) cos(θ)
C( θ
) si
n(θ)
153
17 A simulated experiment; Geiger-Marsden experi-
ment
In nuclear and high energy physics, which deal with probing the structure of very small ob-jects, the tool of exploration is scattering. We see the shapes of objects by observing eitherlight directly emitted by luminous objects, or by seeing light scattered off of the object fromsome other source. In order to see into very small objects, light or other particles of verysmall de’ Broglie wavelength must be used. Fairly long ago the technology to produce beamsof particles such as alpha-particles (Helium nuclei) was developed and used to probe intothe interior of atoms, in order to determine how the atom is constructed. The definitiveexperiments were conducted by Rutherford and Marsden.
Consider a beam of hard sphere particles colliding with a collection of targets, with a fixedtarget density per unit volume. If a projectile can hit a certain area A(θ) around any targetin the entire length L of the target sample, it will be scattered by a scattering angle betweenθ and θ + dθ. If we have an incoming flow of projectiles at rate Rp = dNp
dtdistributed uni-
formly and flowing through an area A, we have a particle beam
AA
L
Ntarget
of flux F = Rp
A. The rate then with which projectiles scatter by angle θ to θ+dθ is determined
by how many will hit the necessary area Ntarget · A(θ). This will be
dN(θ)
dt= Rscatt(θ) = F ·Ntarget · A(θ)
154
AA
Ntarget
The area A(θ) is called the differential cross section, and is an annular region within theconfines of a certain impact parameter b and b+db, so that A(θ) = 2πb db, which we haveprojected onto the targets in the two figures above.Consider classical scattering of particles of mass m, initial speed v0 and impact parameter bby a static central force field. If all particles with impact parameter in the range b→ b+ dbare scattered into solid angle
b
m
d
φθ
θ
Ω → Ω + dΩ. What is the meaning of the so called scattering cross section? Itrepresents the size of the target that an incoming projectile must hit in order to be scatteredthrough an angle between θ and θ + dθ. For a given target then if a projectile hits the
155
ring of area 2πb db in the figure, it will scatter through such an angle. Suppose that anincoming particle beam has a flux of F particles per second per unit area. The rate withwhich projectiles scatter into θ → θ + dθ is
d2N
dt= F · 2πb db
We translate this from a function of b to a function of θ with
d2N
dt= F · 2πb db
dθdθ
and now find the number of particles per unit time scattered into the cone of outer angleθ + dθ and inner angle θ, which subtends a solid angle dΩ = 2π sin θ dθ;
d2N
dt= F · b db
dθ
1
sin θ(2π sin θ dθ) = F · b
sin θ
db
dθdΩ
and finally we define that quantity loved by particle physicists; the differential cross section
dσ(θ)
dΩ=
1
Fd2N
dt dΩ=
b
sin θ
db
dθ
The trajectory of the particle can be gotten from the energy expression
E =m
2(r2 + r2φ2) + V (r)
but we can eliminate t in favor of φ using the angular momentum equation
` = mr2dφ
dt
or by inverting
d
dt=
`
mr2
d
dφ
We find that
E =`2
2mr4(dr
dφ)2 +
`2
2mr2+ V (R)
or
dφ =`dr
mr2√
2m
(E − V (r)− `2
mr2 )
Now set u = 1r
and
φ1 − φ0 =∫ u1
u0
−du√
2mE`2− 2mV
`2− u2
156
If we let the initial and final values of r be ∞ and the point of closest approach be r0, thenwe find that φ1 = π, φ0 = θ + ψ and π = 2ψ + θ and
θ = π − 2∫ umax
0
bdu√
1− V (u)E− b2u2
with the angular momentum given by
` = mbv0 = b√
2mE
We now specialize to the case of the coulombic potential V (u) = ku and compute the pointof closest approach
r0 =1
umax
The radial speed is zero at this point and so r = 0 and
E = kumax +l2u2
max
2m
umax =−k +
√k2 + 4b2E2
2b2EWe can now compute the integral for the scattering angle and obtain
θ = π − 2 sin−1(1)− 2 sin−1(k√
k2 + 4b2E2)
which can be solved for b in terms of θ
b =k
2Ecot(
θ
2)
which in turn can be inserted into the cross section formula
dσ(θ)
dΩ=
k2
16E2csc4(
θ
a)
An alternative approach that does not involve integral transformation is to use the orbitalequation to find the angle between the asymptotes of the hyperbolic path taken by the pro-jectile.
For Coulombic scattering from a nucleus, V (r) = kr
with k > 0,
E =1
2m((
d
dtr)2 + r2(
`
mr2)2) +
k
r
the point of closest approach is,
1
rmin
=1
umax
= −mk`2
+
√
(mk
`2)2 +
2mE
`2
We know that paths will be hyperbolas, such as in the figure
157
x
y
φ0
Θ
umax
u =1
r= A−B cos θ
with asymptotes at θ = ±φ0 so use conservation of angular momentum
d
dt=
`
mr2
d
dθ
to write
E =1
2m((
d
dtr)2 + (r
d
dtθ)2) +
k
r=
1
2m((
`
mr2
dr
dθ)2 +
`2
m2r2) +
k
r
again use u = 1r
to get
E =1
2m((
`
m
du
dθ)2 +
`2
m2u2) + ku
Insert our hyperbola solution to get
E =`2
2mB2(1− cos2 θ) +
`2
2m(A2 − 2AB cos θ +B2 cos2 θ) + k(A− B cos θ)
The coefficient of cos θ must be zero;
0 = −kB − `2
mAB
or
A = −mk`2
and so
B = −√
(mk
`2)2 +
2mE
`2
158
and we have our complete path. To get the asymptotes we note that at a point where r →∞,u→ 0 and so at θ = ±φ
0 = A−B cosφ0
and so
cosφ0 =A
B=
1√
1 + 2E`2
mk2
or
tan2 φ0 =2E`2
mk2
but the scattering angle is related to the asymtote angle by
2φ0 + π = θ
so
` =
√
mk2
2Etanφ0 =
√
mk2
2Ecot
θ
2
Remember that
E =1
2mv2
0, ` = mv0 b
where b is the impact parameter.
The cross section derived above applies to one target. Suppose a beam of projectiles is shotat a target that consists of many nuclei, such as a gold sheet? The number of gold nucleiper unit volume is ρ, and so the number of targets per unit volume is
ntarget =ρNA
M
where M is the atomic mass of gold, and NA is Avagadro’s number. If the target is gold foilof cross-sectional area A and thickness L, we have
Ntarget =ρNA
MAL
gold nuclei, which we multiply our cross section computed earlier by, to arrive at
d2N
dt= F ·Ntarget · b
db
dθ
1
sin θ(2π sin θ dθ)
This quantity can be measured experimentally; if we bombard the target for time T we willhave accumulated
dN = F ·Ntarget · T · bdb
dθ
1
sin θ(2π sin θ dθ)
scattered projectiles in the detector positioned to count projectiles scattered into the coneof solid angle 2π sin θ dθ centered on angle θ.Rutherford, Marsden and Geiger bombarded thin gold foil with alpha particles. They werewatching for substantial back scattering at angles close to 180 degrees, since the plum pud-ding model lacked a hard, localized positive scattering center there should be practically
159
no backscattered alphas. They found instead that 1 in 10,000 alphas scatter through an-gles exceeding 90 degrees, rather than the 1 in e(90)2 predicted by the plum pudding model.Rutherford himself was surprised and compared it to firing a 15 inch naval artillary shellat a sheet of tissue and having the shell bounce back at you. To explain this required allof the positive charge in the atom to be localized at a single very massive point with theelectrons orbiting in a cloud in sufficient numbers to make the atom neutral. The classicalscattering formula for a pure Coulombic nuclear potential gives very good agreement withthe experimental data.The details of this very thorough and brilliantly executed experiment are quite remarkableconsidering the era in which it was performed. The α source used, a sample of purifiedradium, was very potent by modern standards; a butt-kicking 0.1 Curie. This is enough toemit several billion α particles per second in the decay process
222Rn→218 Po+ α
The detector used was a micrometer-mounted zinc-sulfide screen that would briefly flashwhen an α struck it. The screen could be moved from one angular position to another, andwas visually monitored with a microscope.The experimenters studied the scattering rate as a function of incoming α particle energy,scattering angle, and several other factors. Rutherford analyzed the data and confirmed thatit supported the hypothesis that the atom possesses a concentrated and very small nucleus.
Bombardment of a thin film of gold, such as that used by Rutherford, Geiger and Mars-den, with α particles (4
2He nuclei) requires an apparatus very similar to that used in theBragg scattering experiment. Chemists often determine the lattice spacings of crystallinesamples using a powder diffraction apparatus, in which a fine powder of the crystal isbombarded with X-rays, essentially an X-ray version of Bragg scattering. The sample is heldat the center of a flat cylindrical disk container with a removable top plate. The entire innerwall is lined with photographic film, and X-rays enter through a port. The scattered X-raysexpose the film, and by careful measurements of the positions of exposed areas, the angulardisplacements of diffraction maxima can be found.
160
R
s s
s=R θ
Photographic film liner
entryport
sample
Scattering of electrons requires more sophistication, such as in the spherical detector illus-trated below, which uses perhaps hundreds of scintillation detectors to count the scatteredelectrons. Two colliding beams scatter from the center point of the machine, and the elec-tronics of the scintillation detectors count events and report them versus the scattering anglecosine.
161
0 50 100 150100
101
102
103
104
105
counter
coun
ts
1
2
34
5
6
7
1’
2’
3’4’
5’
6’
7’
θ
Sum to getcounter 1signal
Six million particles bombarding total α particles incident on a stationary Au nucleus at thecenter of the detector produces the following plot of the number of counts dN gotten by thedetectors located at angle θ, versus cos θ. Out of 6.0× 106 incident particles (total counts),there were 1976 particles scattered through angles with θ > π
2= 90o, 3 particles scattered
at angles exceeding 169o! If the gold atom was well described by the Plum-Pudding Model
162
of Thompsen, this would be completely impossible.
−1.0 −0.5 0.0 0.5 1.0100
101
102
103
104
105
cos θ
dN (
Ω)
Plotted against this is the theoretical curve obtained from the assumption that there is anucleus, namely
dN(θ) = F b
sin θ
db
dθdΩT Ntarget
= 2πF b dbdθ
dθ T Ntarget
in which T is the time of exposure (the time for which the sample was bombarded). Thismakes F T equal to the total number of projectiles (6.0× 106) divided by the cross sectionalarea of the beam.You can see that the data conforms extremely well with the hypothesis of a concentratednucleus.In the simulated experiment below you will be asked to analyze a similar data set, and com-pare it to the theoretical predicted dN(θ) for several models of the atom.
17.1 The numerical experiment
The figure illustrated above is the data analysis of a simulated scattering experiment. TheRutherford-Geiger-Marsden experiment is a bit too costly to permit its performance on our
163
campus, but we can certainly analyze a generated data set.
Apparatus. We imagine that an apparatus utilizing scintillators such as that on the pre-ceeding page was used, with a detector density of one scintillator per degree in the θ direction,making dθ = π
180.
The incident beam. In this simulation, my incident beam was made by placing a purifiedradium sample of 1.0 × 10−4 Curies behind a series of collimating aperatures to cut downthe the isotropically emitted particles into a tight beam. The α particles are found to havea kinetic energy of 113, 760 eV , making them deeply nonrelativistic.
Collimators
Lead shield
Radium sample
The runtime of the experiment is T = 60.0 seconds, during which time the scintillators record6.0× 106 events (incident α particles), and so Np = 6.0× 106. The machine reports countsin a manner similar to the operation of a scaler-counter; each event triggers the scintillatorthat detects it to report its number (1→ 180) to a computer, which writes the number to afile. Counter n has therefore has angular location θ = n π
180in radians. The data file must
then be analyzed to create a histogram or frequency plot of counts versus cosine of the angleat which the event was detected.The target. The target is a very thin gold foil of thickness L = 4.25 × 10−10m. This isimpractically thin, about ten close-packed atoms deep, but manufacturing techniques haveevolved considerably since the time of Geiger and Marsden. Gold layers a few atoms thickcan be laid down on a substrate by sputtering or ballistic deposition. The density of gold is1.93 × 104 kg
m3 , and its atomic mass is 197 g
mole. Since the gold atoms are close-packed, and
have a radius of about 1 × 10−10m, no individual α is likely to interact with a gold atomwith an impact parameter exceeding 1.0× 10−10m.
17.2 Raw data
The actual data is given in the table below. Each scintillator array for angle θ = n π180
iscounted by an MCA, which tabulates the counts and reports them by channel n. There isas usual a linear relation between channel n and in this case the angle θ.
164
17.3 Problem 1. Analysis of simulated lab data
You should now work up this data, and determine which hypothesis it conforms to; a nuclearatom, or a Thompson model atom. To do so you must first plot the data as dNscatt(θ), thenumber of projectiles scattered into a cone of angle θ to θ + dθ, versus the cosine of thescattering angle. You should use a log plot on the y-axis because of the tremendous variationin dNscatt with angle.Next you should use the information given regarding beam energy, target composition andthickness, and scattering time, to find each of the following items;
F ·Ntarget =Rp
A· ρNAAL
M
Notice that if we integrate over all scattering angles, since every incident α is scattered bysome angle, Nscatt =
∫ π0 dNscatt(θ) dθ;
Nscatt = Np = RpT = πb2maxF ·NtargetT = πb2max
Np
AT· ρNAAL
MT
This should be used to verify the supposed value of bmax. You will also need
bdb
dθ, for the Rutherford model
dθ, the angular separation of scintillators
Next you will need to compute the theoretical number of projectiles scattered into eachscintillator using
dNscatt(θ) = 2πF b dbdθ
dθ T Ntarget
for the Rutherford nuclear model, plot the experimental data against the theoretical curveof dN(θ) versus cos θ.You should answer the questions below as well.
1. Explain how you can measure the energy of the α particles used in the simulated experi-ment.
2. Compute the distance of closest approach to a gold nucleus of one of the α particles usedin the simulated experiment in a head on collision b = 0.
3. Compute the α particle beam rate Rp.
165
Eα = 113760 eVChannel Counts Channel Counts Channel Counts Channel Counts
0 474473 42 160 85 16 127 61 3978818 43 153 86 12 128 62 858939 44 137 87 14 129 53 301879 45 137 88 15 130 14 138943 46 139 89 13 131 35 75729 47 105 90 17 132 66 45606 48 126 91 10 133 47 29514 49 96 92 14 134 58 20396 50 116 93 13 135 29 14383 51 81 94 16 136 510 10634 52 91 95 10 137 511 8093 53 72 96 11 138 312 6476 54 77 97 10 139 313 4933 55 73 98 11 140 814 3947 56 76 99 16 141 315 3385 57 70 100 8 142 316 2747 58 65 101 4 143 317 2313 59 50 102 7 144 218 1902 60 52 103 12 145 319 1634 61 47 104 4 146 320 1466 62 53 105 10 147 221 1242 63 51 106 7 148 522 1077 64 34 107 11 149 323 989 65 46 108 9 150 124 893 66 42 109 10 152 225 754 67 29 110 5 153 126 661 68 44 111 6 154 227 584 69 38 112 8 155 128 558 70 36 113 6 156 229 464 71 35 114 7 157 130 432 72 34 115 5 158 131 400 73 42 116 5 159 132 357 74 30 117 7 160 233 314 75 35 118 8 161 134 292 76 29 119 2 163 135 297 77 34 120 6 164 236 281 78 24 121 7 165 237 253 79 26 122 5 166 138 222 80 23 123 5 167 139 184 81 31 124 3 169 140 179 82 21 125 9 170 141 165 83 18 126 5 171 1
166
17.4 The Thompson model
Demonstrate using Gauss’s law that within the Thompson atom, the force exerted on an αparticle due to the positive charge alone is
F =2Ze2
4πε0R3(xi + yj)
θ
θ
bR
R
The α simply blows the electrons, 18000
times its mass, away, and so interacts mostly withthe positively charged part of the atom. If the incident α with impact parameter b < R andvelocity v = −v0 i enters the atom at point r(0) = −
√R2 − b2 i+ b j, it will be pushed away
by the force above. Show that the solutions to the Newtonian equations of motion are
r(t) = (v0√γ
sinh√γ t−
√R2 − b2 cosh
√γ t)i + b cosh
√γ t j
in which
γ =2Ze2
4πε0mαR3
giving the position vector of the α after entering the atom. This is only valid as long as|r(t)| ≤ R, since after the α leaves the atom, it experiences no force, and so travels in astraight line.Prove that the time at which the α leaves the atom is
tanh(√γ t`) =
2v0
√R2 − b2
√γ(R2 +
v20
γ)
and demonstrate that the scattering angle is given by
tan θ =vy(t`)
vx(t`)=
2bγ√R2 − b2
v20 − γ(R2 − 2b2)
167
Demonstrate that the electrons embedded in the atom will execute elliptic orbits, with thecenters of the ellipses being at the center of the atom, rather than the focal points being atthe center of the atom, as is the case in planetary motion.The final phase is to establish a reasonable value for R. It was thought in 1912 that theatom was about R = 1.0× 10−10m in radius.If we suppose that R = 1.0× 10−10m, then for gold (Z = 79)
Ze2
4πε0R= 1137.6 eV
Geiger and Marsden used α particles of about 5.5MeV , which is very high. In our simulatedexperiments we will use much lower energies.Show that for the atomic size of R = 1.0× 10−10m,
tan θ =2 b
R
√
1− b2
R2
Eα
1137.6 eV− (1− 2b2
R2 )
For the α energy of 113760 eV , compute the de’Broglie wavelength. You can see that this iscertainly small enough to reveal details within the atom, given its estimated size.
17.5 Monte Carlo scattering simulation
One of the most commonly performed numerical experiments today in high energy physics isthe Monte Carlo simulation of scattering. Runtime on big experimental hardware is expen-sive, so experiments are first designed by simulation before hardware is built or modified, andruntime is paid for. Typically one generates scattering events that obey the dynamics of sometheoretical model, and precise projectile parameters such as impact parameter are simulatedusing random number generation. You will now run a Monte Carlo for the Thompson model.
The actual Monte Carlo portion of the simulation is the beam generation. To simulate abeam of spatially uniform flux, consider a phosphorescent screen placed perpendicular to thebeam. Projectile impacts should be randomly and uniformly distributed, as illustrated below.
168
−1.0 −0.5 0.0 0.5 1.0−1.0
−0.5
0.0
0.5
1.0
This uniform beam was simulated by choosing two randomly and uniformly distributednumbers between −1 and 1, one to act as x, one to act as y. The PDF’s of these numbersare both 1
2. The resulting impact parameter b =
√x2 + y2 is randomly distributed, but not
uniformly;
℘[b ≤ a] =∫
b≤a
dx dy
4=∫ a
0
b db dθ
4=πa2
8
and so b has an unnormalized PDF of
fb(b) =d
db′℘[b′ ≤ b] =
πb
4
Suppose that bR
is a randomly distributed variable on the interval from 0 to 1 with such aPDF. What is the probability that an α will be scattered by an angle greater than π
2radians?
This would require that the denominator be negative.
Write a simple computer program to generate 600, 000 scattering events with such a randombeam of α particles.1. Generate a uniform random deviate r, interpreted as b
R, with a normalized PDF based
on that given above. The easiest way would be to generate two random deviates x and y,both between −1 and 1, and if r =
√x2 + y2 ≤ 1, proceed, otherwise pick two more and try
again. You could use the inverse method or the accept-reject algorithm instead.2. Compute the appropriate scattering angle θ from the Thompson scattering formula. Dothis for several α particle energies.3. Compute which scintillator would detect the scattered α from n = floor(180 · θ
π) where
floor is the function that rounds a floating point number down to the nearest integer. Printout this integer.
169
4. Now that you have a data file of detector events, create a frequency plot of the numberof counts gotten by each detector. Does this look anything like that seen by Geiger andMarsden?
17.6 Energy dependence in the Geiger-Marsden experiment
Geiger and Marsden were very thorough. They also investigated the dependence of thescattering rate on incident α particle energy.Beginning with
dN(θ) = 2πF Ntarget Tb db
and
b =k
2Eα
cotθ
2
we can determine the energy dependence of the scattering rate on incident energy by countingthe number of α particles that should be scattered into angles exceeding π
2radians. These
must be incident on gold nuclei with impact parameters less than
bmax =k
2Eα
cotπ
4=
k
2Eα
and so the total number of scattering events logged by all detectors located at θ > π2
shouldbe
Nθ> π2
=∫ bmax
02πF Ntarget Tb db = πF Ntarget Tb
2max = πF Ntarget T
k2
4E2α
This inverse-energy-squared behaviour is almost universal in high energy physics scatteringexperiments. In this problem you will verify the hypothesis that the scattering rate has thisbehaviour as a function of incident energy.
On the next few pages you will find the detector counts for several incident α particle en-ergies. The data for Eα = 113760 eV was displayed several pages back. The radium sourceemits α particles with a variety of energies. One energy can be singled out by passing thebeam through a velocity selector; crossed electric and magnetic fields. This sub-beam canthen be accelerated further with a high voltage.
For each energy for which you have the scattering data, determine the number of α parti-cles scattered through more than π
2radians. Plot this versus the energy. Now compute the
number of such events predicted by the Rutherford formula, given above, four our beam andtarget parameters. In each experiment, we have logged 6.0× 106 scattering events.Plot the experimental counts and the theoretical curve on the same figure, compute χ2 forthe four data points. Is the hypothesis verified to be true?
170
Eα = 227520 eVChannel Counts Channel Counts Channel Counts Channel Counts
0 4453047 35 67 71 6 109 31 1160804 36 61 72 6 110 22 214677 37 55 73 6 111 33 75138 38 60 74 8 112 14 34747 39 54 75 5 113 25 18774 40 49 76 6 114 26 11402 41 50 77 4 115 17 7330 42 52 78 4 116 28 5061 43 42 79 9 117 19 3524 44 35 80 4 118 210 2701 45 39 81 3 119 211 2098 46 35 82 5 120 212 1626 47 25 83 7 121 313 1249 48 28 84 2 122 114 1017 49 21 85 8 123 315 828 50 18 86 3 127 216 692 51 16 87 1 130 117 564 52 18 88 9 131 118 529 53 19 89 4 132 119 438 54 16 90 5 133 120 341 55 15 91 1 134 121 299 56 18 92 10 135 122 274 57 11 93 1 137 123 239 58 15 95 2 138 124 224 59 11 96 4 139 125 207 60 14 97 4 142 226 170 61 16 98 3 143 127 150 62 5 99 4 147 128 122 63 8 100 2 149 229 118 64 7 101 2 151 230 128 65 8 102 4 154 131 99 66 9 103 3 155 132 81 67 10 104 4 159 133 81 68 11 105 1 162 134 70 69 9 106 4 163 1
171
Eα = 56880 eVChannel Counts Channel Counts Channel Counts Channel Counts
1 474793 45 508 89 69 133 162 2776956 46 520 90 58 134 183 1202451 47 493 91 61 135 54 556203 48 433 92 63 136 105 302768 49 406 93 51 137 136 183063 50 368 94 79 138 127 118825 51 368 95 49 139 158 81131 52 334 96 56 140 129 57792 53 324 97 48 141 1310 42979 54 317 98 44 142 1111 32755 55 267 99 45 143 912 25560 56 279 100 44 144 1113 20046 57 263 101 50 145 614 16380 58 238 102 35 146 1315 13175 59 247 103 41 147 616 11103 60 224 104 43 148 1117 9286 61 204 105 39 149 1018 7731 62 195 106 26 150 619 6606 63 183 107 41 151 1120 5644 64 195 108 40 152 821 4953 65 177 109 35 153 922 4323 66 194 110 40 154 623 3802 67 163 111 33 155 724 3499 68 165 112 37 156 825 2979 69 145 113 34 157 726 2590 70 147 114 37 158 827 2342 71 122 115 30 159 1028 2033 72 112 116 36 160 529 1993 73 125 117 29 161 530 1748 74 112 118 31 162 631 1514 75 113 119 31 163 532 1500 76 119 120 23 164 633 1301 77 90 121 22 165 234 1221 78 102 122 24 166 135 1069 79 100 123 13 167 436 988 80 95 124 20 168 237 875 81 112 125 17 169 238 906 82 81 126 15 170 339 789 83 82 127 19 171 140 759 84 98 128 15 172 441 698 85 66 129 18 173 242 664 86 99 130 19 174 143 586 87 61 131 13 175 244 596 88 68 132 15 176 1
172
Eα = 28440 eVChannel Counts Channel Counts Channel Counts Channel Counts
3 476108 47 1840 91 268 135 694 1569314 48 1768 92 222 136 525 1210099 49 1597 93 236 137 626 729550 50 1474 94 219 138 507 472849 51 1461 95 216 139 558 324276 52 1320 96 214 140 579 231905 53 1239 97 217 141 5310 172040 54 1176 98 202 142 5611 130800 55 1166 99 175 143 5412 101969 56 1125 100 182 144 5413 81012 57 1044 101 167 145 5314 65076 58 966 102 175 146 5115 53825 59 885 103 171 147 4416 44150 60 924 104 179 148 3717 37007 61 848 105 185 149 3418 31211 62 822 106 151 150 2719 26514 63 742 107 158 151 2920 23014 64 752 108 153 152 3121 20010 65 681 109 137 153 2922 17451 66 617 110 123 154 2823 15256 67 620 111 125 155 2624 13624 68 565 112 120 156 2225 11972 69 617 113 121 157 2226 10545 70 563 114 118 158 2027 9500 71 538 115 122 159 2428 8568 72 540 116 103 160 2229 7665 73 466 117 120 161 1730 6969 74 493 118 103 162 1931 6444 75 460 119 108 163 1332 5712 76 441 120 111 164 2033 5277 77 400 121 88 165 1934 4840 78 444 122 120 166 1335 4427 79 393 123 81 167 1636 3981 80 350 124 98 168 1337 3712 81 389 125 82 169 1638 3411 82 374 126 83 170 1039 3199 83 339 127 75 171 1240 2947 84 322 128 82 172 741 2759 85 303 129 68 173 542 2522 86 272 130 97 174 443 2343 87 320 131 76 175 444 2326 88 260 132 60 176 645 2025 89 260 133 61 177 346 2066 90 266 134 57 178 1
173
17.7 Problems
1. Consider a simplified atomic model in which the positive charge +Ze is spread out intoa uniform ball of radius R, and the Z electrons are embedded in this ball in such a waythat the electrostatic potential outside of the ball is zero, and inside is V0 > 0. Incomingα particles of impact parameter b and speed v0 will slow down once they get into the ball,to speed v1. The angular momentum will be conserved, requiring that the α beam bendsharply at r = R as illustrated below.
b
s
φψ ψ
Θ
Θ
φR
The impact parameter of the bent beam is s.a. If the incoming beam has angle of incidence φ with respect to the normal of the surfaceof the ball, and has kinetic energy exceeding V0, the transmitted beam will be bent to angleλ with respect to the normal. Use conservation of energy and angular momentum togetherwith the figure to show that
sinφ
sinλ=v1
v0
=b
s
b. Use this to find the scattering angle θ as a function of b. Compute the maximum scat-tering angle as a function of m (α particle mass), b, and V0. Perhaps you can see now whyRutherford was so surprised when α particles literally bounced back into his face.
c. Find the scattering angle in the case of 12mv2
0 < V0.
174
2*. For the case 12mv2
0 > V0 in the first problem, find the differential cross section dσdΩ
. Thisis a slightly more challenging problem.
3. For a stream of 6.0MeV alpha particles incident on a gold foil of thickness 1× 10−6m,determine the rate at which particles are scattered through angles greater than 0.1 radian.
17.8 Notes
The data was generated using energies
Eα =k
Rc
with Rc = 0.2× 10−10 meters for the 113760 electron volt α. Maximum impact parameterwas in practicality around 1× 10−10 m. This makes
πFNtargetTk2
4E2α
=
π(6× 106
60 s)(60 s)
(1.93× 104)(6.023× 1023)(4.25× 10−10)
(0.197)
R2c
4= 1.182× 10−26R2
c
which will give 118, 473, 1890, and 7563 α particles scattered by more than π2
radians forRc = 0.01, 0.02, 0.04 0.08 × 10−10m which gives the four energies used to generated theenergy dependence data. Physically counting in each case we find 116, 471, 1885 and 7540,so that the simulations are quit good.
/* Monte Carlo that creates data for Thompson atom scattering */
#include <stdlib.h>
#include <math.h>
#include <stdlib.h>
#define PI 3.1415926
int n;
main()
float m1,m2,test,max,b,theta,R_cl, E,E0,num,den;
int crap,det;
/* seed the random number generator, establish MAX=2^31 */
srand(17);
max=pow(2.0,31.0);
/* constants and the energy */
E=11000.0;
E0=1137.6;
for(n=0;n<600000;n++)
175
/* here is the Monte Carlo step */
generate:
m1=2.0*(float)rand()/max-1.0;
m2=2.0*(float)rand()/max-1.0;
b=sqrt(m1*m1+m2*m2);
if(b>1.0) goto generate;
/* now create the event */
num=2.0*b*sqrt(1.0-b*b);
den=(E/E0)-(1.0-2.0*b*b);
theta=atan2(num,den);
/* print out detector logging event */
det=floor(180.0*theta/PI);
printf("%d\n", det);
Below is the program to generate data for the theoretical Rutherford scattering, using
E =k
0.02× 10−10
sin θ
sin4 θ2
= 2cos θ
2
sin3 θ2
2πbdb
dθdθF T Ntarget
= 2π · ( k
4 k0.02×10−10
)2 1
2cot
θ
2
1
sin2 θ2
· 6× 106 · (1.93× 104)(6.023× 1023)(4.25× 10−10)
(0.197)
π
180
= 2π(0.02× 10−10)2
4 · 4sin θ
sin4 θ2
· 6× 106 · (1.93× 104)(6.023× 1023)(4.25× 10−10)
(0.197)
π
180
= 2π (37.62)sin θ
sin4 θ2
π
180
#include<stdio.h>
#include<math.h>
#define PI 3.1415926
int n;
float dtheta,theta,sig;
main()
dtheta=1.0;
for(n=1;n<180;n++)
theta=(float)n*dtheta*PI/180.0;
sig=sin(theta/2.0);
sig=sig*sig;
176
sig=sig*sig;
sig=2.0*PI*(37.62)*sin(theta)*(PI)/(180.0*sig);
printf("%f\t%f\n", cos(theta),sig);
Summing over all output values, we should come up close to six million. The Monte Carloprogram to create the “experimental data” is below;
/* creates data for rutherford scattering */
#include <stdlib.h>
#include <math.h>
#include <stdlib.h>
#define PI 3.1415926
int n;
main()
float m1,m2,test,max,b,theta,R_cl;
int crap,det;
srand(17);
max=pow(2.0,31.0);
R_cl=0.02;
for(n=0;n<6000000;n++)
generate:
m1=2.0*(float)rand()/max-1.0;
m2=2.0*(float)rand()/max-1.0;
b=sqrt(m1*m1+m2*m2);
if(b>1.0) goto generate;
theta=2.0*atan2(R_cl,2.0*b);
/*printf("%f\n", theta); */
/* modify to print out detector number */
det=floor(180.0*theta/PI);
printf("%d\n", det);
177
18 Appendix
18.1 Appendix I. Reformatting Lab Data with Sed and Awk
Sed and awk are two Unix editting tools that are particularly useful in processing data takenin lab. Awk is programmable and uses the syntax of C and shell programming.If you have little or no programming experience, and it is difficult for you to use the supportprograms included in the manual, try to use Awk to process your data. It is easy to learn,and there are lots of Awk scripts strewn throughout the manual. Awk is a free download forWin32 machines from the physics department website, and is part of any Linux installation.As an example we will process data for the Nuclear Counting Statistics experiment.Suppose that we count nuclear decays for 10 second periods, obtaining the following 20measurements
1016
1207
1186
1244
1110
1099
1099
1185
1220
1286
1117
1280
1190
1083
1177
1189
1200
1201
1188
1291
which we enter into a file called list , one number per line as shown above, so the command
cat list
produces the output above. We need to find the largest and smallest values, easy for 20 datapoints, but what if we had hundreds? We can sort the data using the Unix program sort,with the −n switch telling sort to sort by arithmetic value.
sort -n < list > sorted_list
which looks like this
178
1016
1083
1099
1099
1110
1117
1177
1185
1186
1188
1189
1190
1200
1201
1207
1220
1244
1280
1286
1291
This file will have a first line that is blank, a feature of some versions of sort, this means thatthe second line of the file sorted list and the last lineare the smallest and largest entriesin our data file. Check for this behaviour when running sort for the first time .We can pipethe output of this program into an awk script min max.awk that will read the file andreport the second and last records to the console. We need to tell awk that it will read a filewhose records are single lines in column format, one record per line. The record separatorRS will be a blank line, beginning with a null character, the field separator FS is a newlinecharacter.The normal operation mode of awk is that awk acts on each line of a file in turn, treatingeach line as a record with several data fields, each separated by a field separator symbol thatis user defined. The fields are accessed by refereing to them by the names $1, $2, and soforth. The last data field is called $NF. Here is the awk script min max.awk
#min_max.awk
BEGIN FS = "\n"; RS = ""
print "smallest is " , $2, " " , " largest is", $NF
We can obtain the largest and smallest numbers in our data set by running sort on the filelist and piping the output into this awk script
sort < list | awk -f min_max.awk
which produces the output line
.
Smallest is 1016 largest is 1291
179
The next stage in processing the data for this experiment is to sort the data into bins of agiven width in count-space. The range of counts is
Nmax −Nmin = 1291− 1016 = 275
We can sort our data into 11 bins of width 25 counts by running the data in list through thefollowing awk script called bin sort.awk
#bin_sort.awk
i=0
while ( i<11 )
if ( $1 >= 1016+25*i && $1 <= 1016+25+25*i )
print i, " ", $1
++i
This will run through the loop for each line of the file list in turn, assign each data point toa bin, and print the bin first, then the data point separated by a blank space. The output of
awk -f bin_sort.awk list
is
0 1016
7 1207
6 1186
9 1244
3 1110
3 1099
3 1099
6 1185
8 1220
10 1286
4 1117
10 1280
6 1190
2 1083
6 1177
6 1189
7 1200
7 1201
6 1188
10 1291
We can sort this data by running
180
awk -f bin_sort.awk list | sort -n
which will produce the sorted output below. Note that sort will perform numeric sortingbased on the first field of each data line.
0 1016
2 1083
3 1099
3 1099
3 1110
4 1117
6 1177
6 1185
6 1186
6 1188
6 1189
6 1190
7 1200
7 1201
7 1207
8 1220
9 1244
10 1280
10 1286
10 1291
We can do even better by telling bin sort.awk to print only the bin number of each datapoint and piping the output through sort into the uniq command that will list the numberof occurances of each bin label in the list, and output two columns, bin population followedby bin number.
#bin_sort.awk
i=0
while ( i<11 )
if ( $1 >= 1016+25*i && $1 <= 1016+25+25*i )
print i
++i
We run the command
awk -f bin_sort.awk list | sort -n | uniq -c
which produces output
181
1 0
1 2
3 3
1 4
6 6
3 7
1 8
1 9
3 10
We can save this data to a file called bin pop with
awk -f bin_sort.awk list | sort -n | uniq -c >bin_pop
and print out the contents of the file in a nice format that can be imported into a LaTeXlab report as a table
awk ’ print $2, "&", $1, "\\\\", "\\hline"’ bin_pop
which produces
0 & 1 \\ \hline
2 & 1 \\ \hline
3 & 3 \\ \hline
4 & 1 \\ \hline
6 & 6 \\ \hline
7 & 3 \\ \hline
8 & 1 \\ \hline
9 & 1 \\ \hline
10 & 3 \\ \hline
We now add a few lines to this and we have our processed data in the form of a nice LaTeXtable.
\begintabular|c|c|\hline
Bin label & Bin population \\
\hline
0 & 1 \\ \hline
2 & 1 \\ \hline
3 & 3 \\ \hline
4 & 1 \\ \hline
6 & 6 \\ \hline
7 & 3 \\ \hline
8 & 1 \\ \hline
9 & 1 \\ \hline
10 & 3 \\ \hline
\endtabular
182
and this prints as seen belowBin label Bin population
0 12 13 34 16 67 38 19 110 3
We can include another column for the center of the bin with
awk -f table.bin bin_pop
using the script
#table.awk
print $2, "&", $1, "&", 1016+12+25*$2, "\\\\", "\\hline"
which produces the output
0 & 1 & 1028 \\ \hline
2 & 1 & 1078 \\ \hline
3 & 3 & 1103 \\ \hline
4 & 1 & 1128 \\ \hline
6 & 6 & 1178 \\ \hline
7 & 3 & 1203 \\ \hline
8 & 1 & 1228 \\ \hline
9 & 1 & 1253 \\ \hline
10 & 3 & 1278 \\ \hline
After adding afew LaTeX lines this becomes the table
Bin Bin pop. Bin center0 1 10282 1 10783 3 11034 1 11286 6 11787 3 12038 1 12289 1 125310 3 1278
Awk and sed can not only be used to easily process huge data files for analysis, but also toprepare data processed into convenient reports.
183
18.2 Appendix II; Nonlinear Least Squares error matrix method
Suppose that we have a set of data points with average values and standard deviations
xi ± σi |I = 1, 2, · · ·N
taken in the lab. These are graphed with error bars versus a parameter upon which theydepend, for example consider scattering data, in which the number of counts gotten by ascintillation counter placed at position θ1 is x1(θ1)± σ1. Suppose that the theoretical countnumber versus θ is a function
N(θ) = a0 + a1 cos θ + a2 cos2 θ
or that we wish to fit our data to such a theoretical curve, and must find the coefficients ai.This can be done by finding the best fit to such a function using a minimization process,called nonlinear least squares regression.
Our experimental data is of the form
xi(θi)± σi
The theoretical values corresponding to these experimental values will be
ξi = a0 + a1 cos θi + a2 cos2 θi
and in general have the form
ξi =n∑
m=1
Cimam
if there are n constants ai involved in the theoretical curve expression.
Suppose that the likelihood of xi being the measured value is given by a Gaussian distribution
L =1
√
2πσ2i
e− (xi−ξi)
2
2σ2i
then the probability that the entire set of experimental data actually being found to havethe values gotten in the experiment will be
L =1
(2π)N2 σ1 · · ·σN
e−∑N
i=1
(xi−ξi)2
2σi
We can maximize this probability with respect to the parameters ai, which will compute theparameters describing the theoretical form most closely agreeing with our data out of allpossible theoretical curves in the same family.
Maximizing with respect to the parameters ai leads to the equations
∑
i
(xi − ξi)σ2
i
∂ξi∂am
= 0
184
or∑
i
Cim
σ2i
xi =∑
i,l
CimCil
σ2i
al
We can define the vector of experimental results
Xm =∑
i
Cim
σ2i
xi
and the error matrix
Mml =∑
i
CimCil
σ2i
and then we findX = Ma
which has solutiona = M−1X
Suppose that hundreds of identical experiments have given us the opportunity to compute theset of parameters a many times, and we obtain a set of average values for these parametersa, What is the standard deviation in the values of these fit parameters?
(am − am) =∑
k
(M−1)mk(Xk − Xk) =∑
kj
(M−1)mk
Cjk
σ2j
(xj − xj)
however if we assume that the measurements resulting in each data point are uncorrelated,then
< (xj − xj)(xi − xi) >= σ2i δij
and so
< (am − am)(al − al) >=∑
ijkp
(M−1)mk
Cjk
σ2j
(M−1)lp
Cip
σ2i
σ2i δij
=∑
kp
(M−1)mk(M−1)lpMpk = (M−1)ml
and so the standard deviation in am is
σam=
1√
(M−1)mm
So by finding the matrix M we essentially find the curve of best fit and its tolerances.
ExampleConsider the following data table
185
x f(x) σ0 1.1 0.1
1 2.9 0.1
2 8.9 0.1
3 19.1 0.1
We propose a theoretical formf(x) = a0 + a1x
2
The theoretical values of f(x) should be then
ξ1 = a0
ξ2 = a0 + a1
ξ3 = a0 + 4a1
andξ4 = a0 + 9a1
We find the matrix Cim is
Cim =
1 01 11 41 9
the data vector elements are
X1 =1.1
0.01+
2.9
0.01+
8.9
0.01+
19.1
0.01= 3200
X2 =2.9
0.01+ 4
8.9
0.01+ 9
19.1
0.01= 21040
and the error matrix is
Mlm =
(
100 100 100 1000 100 400 900
)
1 01 11 41 9
=
(
400 14001400 9800
)
This has inverse
(M−1)lm =
(
0.005 −0.0007143−0.0007143 0.0002041
)
186
We arrive at the following parameters(
a1
a2
)
=
(
0.005 −0.0007143−0.0007143 0.0002041
)(
320021040
)
=
(
0.97112.0085
)
and so the functionf(x) = 0.9711 + 2.0085x2
best fits our lab data. Furthermore we know
σa1 =√
0.005 = 0.071, σa2 =√
0.000204 = 0.01428
are the errors in the determination of these best fit parameters.
18.3 Appendix III. The Plotting of Lab Data with GNU plotutils
This suite of graphics utilities can be used to build publication quality graphs from raw data.It consists of the command-line programs graph, plot, spline, ode and several others. Theuse of graph is the subject of this section.Graph can output is work into a variety of formats and devices, chosen with the command-line option -T. For example
graph -T X data
will output the resulting figure into xplot, an X windows primitive graphing canvas. Tooutput to pnm, xfig, Postscript, or GIF, one just passes this choice to graph at invocationand redirects the resulting data into a file, for example
graph -T ps data > data.ps
Read through the voluminous /usr/local/info/plotutils.info file for more command line op-tions, such as those needed to limit the abcissa and ordinates xmin ≤ x ≤ xmax, ymin ≤y ≤ ymax
graph -T ps -x xmin xmax -y ymin ymax data > data.ps
or plotting several data files on the same figure
graph -T ps data1 data2 data3 > data.ps
or perhaps using a logarithmic x-axis within domain 100 ≤ x ≤ 104
graph -T ps -l x -x 0 4 data > data.ps
One can add labels to the x and y axes as well as a banner label
graph -T ps -X ‘‘ x, m’’ -Y ‘‘Volts’’ -L ‘‘Voltage per meter’’ data > data.ps
The labels can have Greek letters or math symbols, which are given later in a table in thisdocument
187
graph -T ps -L ‘‘ \*Q\sp*\ep(x,t)\*Q(x,t) for t=0.1 s’’ data > data.ps
will print “Ψ∗(x, t)Ψ(x, t)fort = 0.1s” along the top of the figure.Graph has the capacity to produce graphs within graphs, or to make multigraphs consistingof many repositioned plots. For example the command
graph -T ps --reposition 0.0 0.6 0.3 -L "a=0.07" 07 --reposition 0.3 0.6 0.3
-L "a=0.09" 09 --reposition 0.6 0.6 0.3 -L "a=0.11" 11 --reposition 0.0 0.3
0.3 -L "a=0.13" 13 --reposition 0.3 0.3 0.3 -L "a=0.15" 15 --reposition 0.6
0.3 0.3 -L "a=0.17" 17 --reposition 0.0 0.0 0.3 -L "a=0.19" 19 --reposition
0.3 0.0 0.3 -L "a=0.21" 21 --reposition 0.6 0.0 0.3 -L "a=0.23" 23 >page1.ps
will create the following multiplot
a=0.07
−15 −10 −5 0 5 10 150
100
200
300
400
500
600
700
a=0.09
−15 −10 −5 0 5 10 150
100
200
300
400
500
600
a=0.11
−10 −5 0 5 100
100
200
300
400
500
a=0.13
−8 −6 −4 −2 0 2 4 6 80
100
200
300
400
500
a=0.15
−8 −6 −4 −2 0 2 4 6 80
100
200
300
400
500
a=0.17
−6 −4 −2 0 2 4 60
100
200
300
400
a=0.19
−6 −4 −2 0 2 4 60
100
200
300
400
500
a=0.21
−6 −4 −2 0 2 4 60
100
200
300
400
a=0.23
−6 −4 −2 0 2 4 60
100
200
300
400
The fact that these programs are command-line means that they can be used in shell-scriptsor batch files to perform bulk processing of graphics files.For plotting discrete data points the switch −S allows you to use a large variety of symbolsfor individual data points
188
Symbol number Symbol1 ·2 +3 *4 ©5 ×
with 31 different symbols in all. For example
graph -T ps -S 4 0.01 data > data.ps
uses circles whose size is 0.01 times the size of the box within which the entire plot is drawn.Plots can be rotated by 90, 180, 270 degrees with –rotation
graph -T ps --rotation 90 > data.ps
Lines connecting data points in plots can be manipulated with -m, line mode 0 is no con-necting line. Line style, width, and even color can be manipulated. For example we use-m to change linemode from solid to dotted or dashed, but adding -C changes this to colorlinemode rather than breaking the line in different ways. We can place a plot within a plotby repositioning the second plot within the first and shrinking it;
graph -T ps -C -m 2 data1 --reposition 0.5 0.5 0.3 data2>data.ps
results in
0 1 2 3 4 5 6−1.0
−0.5
0.0
0.5
1.0
0 1 2 3 4 5 60
1
2
3
4
5
or we can try
graph -T ps -C -m 3 data1 --reposition 0.5 0.5 0.1 data2>data.ps
which produces
189
0 1 2 3 4 5 6−1.0
−0.5
0.0
0.5
1.0
0 1 2 3 4 5 60
1
2
3
4
5
The –reposition is relative to the point (0.0, 0.0) being the lower left corner of the physicaldisplay, and (1.0, 1.0) being the upper right corner. In the first example a miniature virtualdisplay of size 0.3 will be place with its lower left corner at (0.5, 0.5).
These few examples are enough to get started with in using graph, and you really shouldread through the info file to learn more since this barely scratches the surface.One of the most useful features of libplot is the generous collection of font symbols thatit recognizes for drawing strings on plots. These include Greek letters and mathematicalsymbols. For example
graph -T X -L "P\sb2\eb(cos\*h)"
will result in the string P2(cos θ) being drawn on the plot title or label. A partial table ofuseable symbols is given below
symbol escape seq. symbol escape seq. symbol escape seq. symbol escape seq.α \ ∗ a β \ ∗ b Γ \ ∗G χ \ ∗ x∆ \ ∗D ε \ ∗ e Φ \ ∗ F Λ \ ∗ Lµ \ ∗m ν \ ∗ n η \ ∗ y Π \ ∗ PΘ \ ∗H ρ \ ∗R Σ \ ∗ S τ \ ∗ tΥ \ ∗ U Ω \ ∗W Ξ \ ∗ C Ψ \ ∗Qζ \ ∗ z δ \ ∗ d γ \ ∗ g σ \ ∗ sθ \ ∗ h π \ ∗ p φ \ ∗ f λ \ ∗ lψ \ ∗ q ∞ \if ← \ < − → \− >± \+− ∝ \pt ∂ \pd ∩ \ca∪ \cu ∇ \gr ∑ \SU ∫ \is∮ \li h \hb † \dg ℘ \wp∅ \es ⊕ \c+ ⊗ \c∗ ℵ \Ah6= \! = ≥ \ >= ≤ \ <=
√ \sr
190
Some of these symbols are only available if pl fontname has been used to set the font toHershey or HersheySans. Subscripts and superscripts can be created using the delimitersillustrated by the following example;
graph -T X -X"\*Q\sp*\ep\\sbn,l,m\eb(r,\*f,\\*h)"
writes Ψ∗n,l,m(r, φ, θ) to the x-axis label,
graph -T ps -L" L\sp\*a\ep\sb n\eb(r)" > out.ps
writes Lαn(r) to the label. This PostScript output is being redirected to a file.
18.4 Appendix IV. Preparing a lab report with LaTeX
LaTeX is very easy to learn and produces documents of superior quality. Nearly every sci-entific textbook and journal is typeset in TeX or LaTeX. LaTeX has the further advantageof being platform independent. In addition it is free software available in binary format forall existing operating systems. It is installed on several of our lab computers. The supportsoftware for this manual contains templates for all of the labs, and the data analysis pro-grams will output LaTeX tables and formatted equations.
LaTeX is designed to typeset equations and tables, and it excels at these tasks. The basicmath equations are typeset in math mode in which offset, centered equations are sur-rounded with double dollar signs, the math mode delimiters. The language is best learnedby example, and so below you will see the LaTeX commands and corresponding typesetequations for a wide variety of mathematics.
$$ \int_-a^\infty f(x) dx= F(a)$$
∫ ∞
−af(x)dx = F (a)
$$ \oint_C f(z) \over z-z_0 dz = 2 \pi i f(z_0) $$
∮
C
f(z)
z − z0dz = 2πif(z0)
$$ \vecA \times \vecB = \vecC $$
~A× ~B = ~C
$$\vec\nabla \cdot \vecE = \rho \over \epsilon_0 $$
~∇ · ~E =ρ
ε0
191
$$ \gamma \rightarrow \beta \bar\beta $$
γ → ββ
$$ \oint \vecp \cdor d\vecq= 2 \pi n \hbar $$
∮
~p · d~q = 2πnh
$$ \sum_n=0^\infty 1 \over n^s = \zetas $$
∞∑
n=0
1
ns= ζs
$$ \prod_n=1^\infty (1-x^2 \over n^2 \pi^2)=\sin x \over x $$
∞∏
n=1
(1− x2
n2π2) =
sin x
x
$$ \lim_n \rightarrow \infty (1-x \over n)^n = e^-x$$
limn→∞
(1− x
n)n = e−x
$$ P(x)=1 \over \sqrt2 \pi \sigma^2e^-(x-\barx)^2 \over 2\sigma^2$$
P (x) =1√
2πσ2e−
(x−x)2
2σ2
$$ \ln |x| = \int_1^x dy \over y$$
ln |x| =∫ x
1
dy
y
$$ \alpha \beta \gamma \delta \pi \lambda \epsilon \sigma \omega $$
αβγδπλεσω
$$ 1=2 \over \pi[\sin x +1 \over 3\sin 3x +1 \over 5\sin 5x +\cdots] $$
192
1 =2
π[sin x+
1
3sin 3x+
1
5sin 5x+ · · ·]
This shows a pretty full range of LaTeX math typesetting commands.
Below is the basic template for the LaTeX documentstyle article. Type this into a fileto use as a template for all of your lab reports.
\documentclass[12pt]article
\pagenumberingarabic
\usepackage[dvips]graphics
\begindocument
\title Your Title Here
\author
author 1 \\
Department of Physics, University of Wisconsin-Parkside \\
900 Wood Road, Kenosha WI \\
\and
author 2 \\
Department of Physics, University of Wisconsin-Parkside \\
900 Wood Road, Kenosha WI \\
\beginabstract
Here is the very short description of your experiment.
\endabstract
\maketitle
\noindent
\bf Section 1.\\
\\
Describe the equipment used, draw a diagram of the apparatus using
xfig and convert it to a postscript figure. It can be included in
the document with the code line \\
\rotatebox270\scalebox0.5[0.5]\includegraphicsapparatus.ps\\
\noindent
\bf Procedure.\\
\\
Describe the experimental procedure.
193
Be to the point, brief and go step by step.
\noindent
\bf Data.\\
\\
Your data must be in tabular form. There is a handout on using awk
and sed to process your data files and put them into LaTeX table format.
Here is a typical 4 column table including some multicolumn rows.
\begintabular|c|c|c|c|\hline
\multicolumn2|c|\bf Trial & \multicolumn2|c|\bf Trial 2\\
\hline
Bin & Number & Bin & Number \\
\hline
1 & 3 & 5 & 9 \\
\hline
2 & 4 & 6 & 3 \\
\hline
3 & 7 & 7 & 1 \\
\hline
4 & 1 & 8 & 2 \\
\hline
\endtabular
Notice that an ampersand is used to separate
the table fields and a LaTeX newline \\
ends each line in the table.
\noindent
\bf Analysis.\\
\\
This is the section in which you perform sample computations
and analyse your data. Any graphs generated can be included
in this section with
\rotatebox270\scalebox0.5[0.5]\includegraphics../ps/graph.ps\\
It may not be necessary to include the rotate box command here.
This is the place where your error analysis and propagation is performed
and your results compared to the accepted (correct) values.
\noindent
\bf Conclusions.\\
194
\\
In this section draw your conclusions.
\enddocument
To format your lab report, type it up using this model and call it report.tex. RunLaTeX on it from the command line in the directory the file lives in
latex report.tex
If there are no errors, you will be left with a new file called report.dvi. Convert this topostscript with
dvips -o report.ps report.dvi
The result of this is report.ps which can be printed. If there are errors the LaTeX enginewill stop on the offending line and point out the error. It wiil prompt you, respond by typing
X
which exits LaTeX. Open your report in the editor and fix the error. Usual LaTeX errorsare unpaired dollar signs and other simple things, easily remedied.
It is well worth your while to become adept with LaTeX if you plan on pursuing Physicsor any technical field. An excellent book is A Guide to LaTeX2ε by Helmut Kopka andPatrick W. Daly, Addison-Wesley, second edition 1995.
Incidentally, what does the table example in the lab template look like?
Trial Trial 2Bin Number Bin Number1 3 5 92 4 6 33 7 7 14 1 8 2
18.5 Appendix V. The Oscilloscope
This is a device used to analyse electronic waveforms and signals. To measure a signal onedoes the following.1. Apply the signal to the Input terminals of the vertical deflection amplifier.2. Horizontal display is controlled by the triggering sections on the face of the scope. ForExternal triggering, sweep is initiated whenever a preset voltage level is reached, in this caseby an externally supplied source, applied to the Trigger in terminals. For Internal triggering,Trigger signal is the same as the input vertical signal, the wave being observed triggers itself.Line setting triggers from the 60 Hz power line that powers the scope. Auto mode meansthat the scopes internal sawtooth generator is used to initiate the sweep. As this voltage
195
increases, the horizontal trace moves from left to right on the screen until a preset value isreached, then the sweep starts over.In all but Auto modes, the voltage at which the sweep is initiated is set by the Stabilitylevel control, which can select the voltage with a positive or negative slope, and an AC orDC portion of the input signal.3. Horizontal sweep rate can be set with the Sweep time
cmcontrol
4. The y(x) relationship between two applied voltages y(t), x(t) can be determined by set-ting Horizontal sweep to External and applying x(t) to the Horizontal input terminal andy(t) to the Vertical input terminal. This bypasses the sawtooth generator and horizontaldeflection is controlled by x(t) If the two input signals are harmonic with frequencies in arational ratio, a Lissajous figure will be displayed on the screen.
- +
10
5
2 10.5
0.20.1
V/cm
Extsec
msec
Horizontalsweep time/cm
input
Vert. Pos.
Horiz. Pos.
mV
20
10
5
21 0.5
10
5
2
10.5 0.2
0.1
50
20105
10.10.1
Input
VerticalV/cm
Focus Intensity Scale illum.
AC off
-
+
Dc
AC
Stability
Trigger in
Auto
Line Int
Ext
Trigger Mode
5020
1051
secµ
18.6 Appendix VI. Support software
The program below “chi2 table gen.c” will generate χ2 in tabular format extremely quicklyusing the methods of Gaussian quadrature. It is limited to no more than about 25 degreesof freedom.
196
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<plot.h>
float gammaf(float x);
float chi2function( float aa, int DOF);
int DF,n,number;
float high, low,dchi,chi,P;
main(int argc, char *argv[])
if(argc < 5 || argc > 5)
printf("chi2_table_gen low high number degree_of_freedom \n");
printf("low is lower bound, high is upper bound (floats).\n");
printf("number is how many table entries to generate\n");
printf("between low and high.\n");
printf("degree_of_freedom is an integer <= 20.\n");
exit(0);
low=atof(argv[1]);
high=atoi(argv[2]);
number=atoi(argv[3]);
DF=atoi(argv[4]);
if(low<=0.0)
printf("Invalid lower bound\n");
exit(0);
if(high < low)
printf("Invalid upper bound\n");
exit(0);
if(DF<1 || DF > 20)
printf("Invalid number of degrees of freedom\n");
exit(0);
dchi=(high-low)/(float)number;
printf("Generating $\\chi^2$ probabilities $\\1\\over 2^%d\\over 2 \\Gamma(%d\\over 2) \\int_x^\\infty y^%d\\over 2-1 e^-y\\over 2 dy$\n\n",DF,DF,DF);
printf("for $%f\\le x \\le %f$\n\n", low,high);
printf("\\begintabular|c|c|\\hline\n");
printf(" $\\chi^2_d$ & $\\wp(\\chi^2\\ge\\chi^2_d)$\\\\\n");
printf("\\hline\n");
for(n=0;n<=number;n++)
chi=low+(float)n*dchi;
P=chi2function(chi,DF);
printf("%f & %f\\\\\n",chi,P);
197
printf("\\endtabular\n");
The program requires the following (called “chi2.c” ) subroutine to compute the performthe quadrature.
/* computation of chi^2 statistic by Gaussian */
/* quadrature. Uses Laguerre polynomial roots */
/* accuracy is limited to 4-5 decimal places */
/* use GMP lib to improve the accuracy */
#include<stdio.h>
#include<math.h>
#define PI 3.1415926
float gammaf(float x);
float chi2function( float aa, int DOF)
/* a=experimental chi^2, N= degrees of freedom */
/* good up to N=14 but is best well below that point */
int m,p;
float sum,denominator,exponent;
float w[15];
float y[15];
/* the abscissa points */
y[0]=9.3307812017e-2;
y[1]=4.926917403e-1;
y[2]=1.215595412;
y[3]=2.26994953;
y[4]=3.66762272;
y[5]=5.42533663;
y[6]=7.56591623;
y[7]=10.12022856;
y[8]=13.13028248;
y[9]=16.65440771;
y[10]=20.77647890;
y[11]=25.62389423;
y[12]=31.40751917;
y[13]=38.53068331;
y[14]=48.02608557;
/* the weight factors */
w[0]=2.18234886e-1;
w[1]=3.42210178e-1;
w[2]=2.63027578e-1;
198
w[3]=1.26425818e-1;
w[4]=4.02068649e-2;
w[5]=8.56387780e-3;
w[6]=1.21243614e-3;
w[7]=1.11674392e-4;
w[8]=6.45992676e-6;
w[9]=2.22631691e-7;
w[10]=4.22743038e-9;
w[11]=3.92189727e-11;
w[12]=1.45651526e-13;
w[13]=1.48302705e-16;
w[14]=1.60059491e-20;
exponent=(float)DOF/2.0-1.0;
denominator=pow(2.0, exponent)*gammaf((float)DOF/2.0);
sum=0.0;
for(m=0;m<15;m++)
sum=sum+w[m]*pow(2.0*y[m]+aa, exponent );
return(exp(-aa/2.0)*sum/denominator);
Also required is the following rather inefficient program for the gamma function “gammaf.c”
#include<math.h>
#define PI 3.1415926
float gammaf(float x)
/* The Gamma function */
float sum, prefactor;
float b[9],arg[9];
int m;
b[0]=1.00;
b[1]=-0.577191652;
b[2]=0.988205891;
b[3]=-0.897056937;
b[4]=0.918206857;
b[5]=-0.756704078;
b[6]=0.482199394;
b[7]=-0.193527818;
b[8]=0.035868343;
199
sum=1.0;
arg[0]=1.0;
prefactor=1.0;
if(x<0.0) /* this is the only recursive call (1) */
return(PI/(sin(PI*x)*gammaf(1-x)));
if(x>2.0)
do
x=x-1.0;
prefactor=prefactor*x;
while(x>2.0);
if(x>0.0 && x<1.0)
prefactor=1.0/x;
x=x+1.0;
for( m=1;m<=8;m++)
arg[m]=(x-1)*arg[m-1];
sum=sum+b[m]*arg[m];
return(prefactor*sum);
everything can be compiled with the command
gcc chi2.c gammaf.c chi2_table_gen.c -o chi2_table_gen -lm
200
Index
accelerating voltage, 53accept-reject, 13, 175activity, 22adiabatic process, 40alpha particle, 20Aluminum crystal, 62angular correlation, 145angular momentum, 36Awk script, 24
background radiation correction, 24Bargmann-Fock space, 143bcc structure, 63beam-splitter, 37Bequeral, 22beta decay, 20Binomial theorem, 41binsort.c, 16Bohr levels, 89Bragg condition, 55, 63Bragg scattering, 53Bravais lattice, 61bulk modulus, 41
characteristic curve, 6chi squared gen.c, 16Cobalt-60, 148coincidence circuit, 148Compton edge, 127Compton scattering, 126conventional cell, 63Coulomb energy, 36Curie, 22
de Broglie relation, 126de’Broglie wavelength, 56degree of freedom, 47density of states, 111dipole selection rule, 145
EM wave, 111equipartition theorem, 47, 112escape sequences, 190
fcc structure, 63Fermi Golden Rule, 145Fermis’ Golden Rule, 143fig, 187four-momentum, 126Franck-Hertz experiment, 83
Geiger counter, 20Geiger tube, 6, 20GIF, 187gnuplot, 25graph, 187
half-life, 5, 21Helium-Neon laser, 36Hershey fonts, 191high pressure gas tank, 49hypergeometric function, 28
interpolaion, 122ionization potential, 83ionizing radiation, 6isotopes, 5
K capture, 36
lattice spacing, 57, 62Laue condition, 63linemode, 189linetype, 189lissojou figure, 43longitudinal waves, 40
markers, 188Miller indices, 63molecular weight, 42
Neon, 36neutrino, 20Newton’s method, 80Normal distribution, 13nuclear spin, 36null measurement, 16
oscilloscope, 43, 85
201
photoelectric effect, 69Planck apparatus, 69Planck radiation law, 114pnm, 187Poisson distribution, 17Poissonian distribution, 8polarization, 145Postscript, 187primitive cell, 62
radiation, 5radio-isotope generator, 23Rayleigh-Jeans law, 112reposition, 188rotational degrees of freedom, 45Rydberg constant, 89
scaler-counter, 5scaler-timer, 20scattering planes, 53selection rules, 145simulation, 13, 65, 150, 175Sodium, 36space lattice, 61specific heat, 41speed of sound, 41Stefan-Boltzmann law, 114stopping voltage, 71symbols, 190
T versus R program, 122translation vectors, 61
unbiased estimator, 17
wave quation, 111Wein’s law, 114work function, 70
202
