Quantization of Charge, Light, and Quantization of Charge, Light, and

EnergyEnergy

• 1. Quantization of Electric Charge;1. Quantization of Electric Charge;

• 2. Blackbody Radiation;2. Blackbody Radiation;

• 3. The Photoelectric Effect3. The Photoelectric Effect

• 4. X-rays and the Compton Effect.4. X-rays and the Compton Effect.

Three great quantization discoveries:Three great quantization discoveries:

1. Quantization of Electrical Charge1. Quantization of Electrical Charge

2. Quantization of Light Energy2. Quantization of Light Energy

3. Quantization of energy of Oscillating 3. Quantization of energy of Oscillating Mechanical System Mechanical System

Quantization of Electric ChargeQuantization of Electric ChargeEarly measurements of Early measurements of ee and and e/me/m..

The first estimates of the magnitude of electric charges The first estimates of the magnitude of electric charges found in atoms were obtained from Faraday law. found in atoms were obtained from Faraday law.

Faraday passed a direct current through weakly conducting Faraday passed a direct current through weakly conducting solutions and observed the subsequent liberation of the solutions and observed the subsequent liberation of the components of the solution on electrodes. components of the solution on electrodes.

Faraday discovered that the same quantity of electricity, Faraday discovered that the same quantity of electricity, FF, , called latter called latter one faradayone faraday, and equal to about , and equal to about 96,500 C96,500 C, , always decomposed always decomposed 1 gram-ionic weight1 gram-ionic weight of monovalent of monovalent ions. ions.

Quantization of Electric ChargeQuantization of Electric Charge

1F (one faraday) = 96,500C1F (one faraday) = 96,500C1F1F decomposeddecomposed 1gram-ionic weight1gram-ionic weight of of monovalent ions.monovalent ions.ExampleExample: If: If 96,500 C96,500 C pass through a solution ofpass through a solution of NaClNaCl,, 23g23g of of NaNa appears at the cathode andappears at the cathode and 35.5g35.5g of of ClCl appearsappears at the anode.at the anode.

1F =N1F =NAAee - Faradays Law of Electrolysis- Faradays Law of Electrolysis

wherewhere NNAA – Avogadro’s number– Avogadro’s number

ee – minimum amount of charge, that – minimum amount of charge, that was was called an electroncalled an electron

Discovery of electron: Thomson’s ExperimentDiscovery of electron: Thomson’s Experiment..

Many studies of electrical discharges in gases were Many studies of electrical discharges in gases were done in the late 19th century. It was found that the ions done in the late 19th century. It was found that the ions responsible for gaseous conduction carried the same charge as responsible for gaseous conduction carried the same charge as did those in electrolysis. did those in electrolysis.

J.J. ThomsonJ.J. Thomson in 1897 used crossed electric and in 1897 used crossed electric and magnetic fields in his famous experiment to deflect the magnetic fields in his famous experiment to deflect the cathode-rayscathode-rays. .

In this way he verified that In this way he verified that cathode-rays cathode-rays must consist of must consist of charged particles. By measuring the deflection of these charged particles. By measuring the deflection of these particles Thomson showed that all the particles have the same particles Thomson showed that all the particles have the same charge-to-mass ratiocharge-to-mass ratio q/mq/m. He also showed that particles with . He also showed that particles with this this charge-to-mass ratiocharge-to-mass ratio can be obtained using any material can be obtained using any material for a source, which means that this particles, now called for a source, which means that this particles, now called electronselectrons, are a fundamental consistent of all matter. , are a fundamental consistent of all matter.

Thomson’s tube for measuring Thomson’s tube for measuring e/me/m..

Electrons from the cathodeElectrons from the cathode CC pass through the slits atpass through the slits at AA andand BB and strike a phosphorescent screen. The beam can and strike a phosphorescent screen. The beam can be deflected by an electric field between platesbe deflected by an electric field between plates DD and and EE or by or by magnetic field. From measurements of the deflections magnetic field. From measurements of the deflections measured on a scale on the tube’s screen,measured on a scale on the tube’s screen, e/me/m can be can be determined. determined.

Crossed electric and magnetic fields. When a negative Crossed electric and magnetic fields. When a negative particle moves to the right the particle experience a particle moves to the right the particle experience a downward magnetic forcedownward magnetic force FFBB=qvB=qvB and an upward electric and an upward electric forceforce FFEE=qE=qE.. If these forces are balanced, the speed of the If these forces are balanced, the speed of the particle is related to the field strengths by particle is related to the field strengths by

v=E/Bv=E/B

q-_

FFEE=qE=qE

FFBB=qvB=qvB

-

FFEE=F=FBB qE = qvBqE = qvB

Thomson’s ExperimentThomson’s Experiment

In his experiment Thomson adjustedIn his experiment Thomson adjusted EE ┴ ┴ BB so so

that the particles were undeflected. that the particles were undeflected.

This allowed him determine the speed of the This allowed him determine the speed of the particleparticle u =E/Bu =E/B. He then turned. He then turned off theoff the BB field field and measured the deflection of the particles on and measured the deflection of the particles on the screen.the screen.

Deflection of the Electron Beam.Deflection of the Electron Beam.

Deflection of the beam is shown with the top plate positive. Deflection of the beam is shown with the top plate positive. Thomson used up toThomson used up to 200 V200 V between the plates. A magnetic between the plates. A magnetic field was applied perpendicular to the plane of the diagram field was applied perpendicular to the plane of the diagram directed into the page to bend the beam back down to its directed into the page to bend the beam back down to its undeflected position.undeflected position.

With the magnetic field turned off, the beam are deflected by an With the magnetic field turned off, the beam are deflected by an amountamount y=yy=y11+y+y22. . yy11 occurs while the electrons are between the plates,occurs while the electrons are between the plates, yy22 after electrons leave the region between the plates. Letsafter electrons leave the region between the plates. Lets xx11 be the be the horizontal distance across the deflection plates. If the electron moves horizontal distance across the deflection plates. If the electron moves horizontally with speedhorizontally with speed vv00 when it enters the region between the when it enters the region between the plates, the time spent between the plates isplates, the time spent between the plates is tt11=x=x11/v/v00,, and the vertical and the vertical component of velocity when it leaves the plates iscomponent of velocity when it leaves the plates is

wherewhere EEyy is the upward component of electric field. The deflectionis the upward component of electric field. The deflection yy11 is:is:

0

111 v

x

m

qEt

m

qEtav yyyy

2

0

121

211 2

121

21

v

x

m

qEt

m

qEtay yyy

The electron then travels an additional horizontal distanceThe electron then travels an additional horizontal distance xx22 in in

the field-free region from the deflection plates to the screen. the field-free region from the deflection plates to the screen. Since the velocity of the electron is constant in this region, the Since the velocity of the electron is constant in this region, the time to reach the screen istime to reach the screen is tt22=x=x22/v/v0 0 ,, and the additional vertical and the additional vertical

deflection is:deflection is:

0

2

0

122 v

x

v

x

m

qEtvy yy

The total deflection at screen is thereforeThe total deflection at screen is therefore

This equation cab be used to determine the This equation cab be used to determine the charge to-charge to-mass ratiomass ratio q/mq/m from measured deflectionfrom measured deflection yy..

21

21

20

2120

212

021 22

1 xxx

v

E

m

qxx

mv

qEx

mv

qEyyy yyy

ExampleExampleElectrons pass undeflected through the plates of Thomson’s Electrons pass undeflected through the plates of Thomson’s apparatus when the electric field isapparatus when the electric field is 3000 V/m3000 V/m and there is a and there is a crossed magnetic field ofcrossed magnetic field of 1.40 G1.40 G. . If the plates areIf the plates are 4-cm4-cm long long and the ends of the plates areand the ends of the plates are 30 cm30 cm from the screen, find from the screen, find the deflection on the screen when the magnetic field is the deflection on the screen when the magnetic field is turned off. turned off.

mmee=9.11 x 10=9.11 x 10-31-31kg; kg;

q=e=-1.6 x 10q=e=-1.6 x 10-19-19CC

The Mass SpectrometerThe Mass Spectrometer

The The mass spectrometermass spectrometer, first designed by , first designed by Francis Francis William AstonWilliam Aston in 1919, was developed as a means in 1919, was developed as a means of measuring the masses of isotopes. of measuring the masses of isotopes.

Such measurements are important in determining Such measurements are important in determining both the presence of isotopes and their abundance both the presence of isotopes and their abundance in nature. in nature.

For example, natural magnesium has been found For example, natural magnesium has been found to consist of to consist of 78.7 %78.7 % 2424MgMg; ; 10.1%10.1% 2525MgMg; and ; and 11.2%11.2% 2626MgMg. These isotopes have masses in the . These isotopes have masses in the approximate ratioapproximate ratio 24:25:2624:25:26. .

Schematic drawing of a mass spectrometer.Schematic drawing of a mass spectrometer.

Positive ions from an ion source Positive ions from an ion source are accelerated through a are accelerated through a potential differencepotential difference ΔΔVV and enter and enter a uniform magnetic field. a uniform magnetic field.

The magnetic field is out of the The magnetic field is out of the plane of the page as indicated by plane of the page as indicated by the dots.the dots.

The ions are bent into a circular The ions are bent into a circular arc and emerge atarc and emerge at PP22 (a plane of (a plane of

photographic plate or anotherphotographic plate or another ion ion detector). The radius of the circle detector). The radius of the circle is proportional to the mass of the is proportional to the mass of the ionion

The Mass SpectrometerThe Mass SpectrometerIn the ion source positive ions are In the ion source positive ions are

formed by bombarding neutral atoms formed by bombarding neutral atoms withwith XX-rays-rays or a beam of electrons. or a beam of electrons. (Electrons are knock out of the atoms by (Electrons are knock out of the atoms by thethe XX-rays-rays or bombarding electrons). or bombarding electrons). These ions are accelerating by an These ions are accelerating by an electric field and enter a uniform electric field and enter a uniform magnetic field. If the positive ions start magnetic field. If the positive ions start from rest and move through a potential from rest and move through a potential differencedifference ΔΔVV, the ions kinetic energy , the ions kinetic energy when they enter the magnetic field when they enter the magnetic field equals their loss in potential energy,equals their loss in potential energy, q|q|ΔΔV|V|::

Vqmv 22

1

The ions move in a semicircle The ions move in a semicircle of radiusof radius RR. . The velocity of the The velocity of the particle is perpendicular to the particle is perpendicular to the magnetic field. The magnetic magnetic field. The magnetic force provides the centripetal force provides the centripetal forceforce vv22/R/R in circular motion. in circular motion. We will use We will use Newton’s second Newton’s second lowlow to relate the radius of to relate the radius of semicircle to the magnetic semicircle to the magnetic field and the speed of the field and the speed of the particle. If the velocity of the particle. If the velocity of the particle isparticle is vv, the magnitude of , the magnitude of the net force isthe net force is qvBqvB, since, since vv andand B B are perpendicular. are perpendicular.

maF

R

vmqvB

2

qB

mvR

The Mass SpectrometerThe Mass Spectrometer

The speedThe speed vv can be eliminate from equationscan be eliminate from equations

andand

Substituting this forSubstituting this for vv22::

Simplifying and solving for Simplifying and solving for ((m/qm/q):):

Vqmv 22

1

qB

mvR

2

2222

m

BqRv

Vqm

BqRm

2

222

2

1

V

RB

q

m

2

22

Separating Isotopes of NickelSeparating Isotopes of Nickel

• AA 5858NiNi ion chargeion charge ++ee and massand mass 9.62 x 109.62 x 10-26-26kgkg is is accelerated through a potential drop ofaccelerated through a potential drop of 3 kV3 kV and and deflected in a magnetic field ofdeflected in a magnetic field of 0.12T0.12T..

• (a) Find the radius of curvature of the orbit of the (a) Find the radius of curvature of the orbit of the ion.ion.

• (b) Find the difference in the radii of curvature of(b) Find the difference in the radii of curvature of 5858NiNi ions andions and 6060NiNi ions. (assume that the mass ions. (assume that the mass ratio isratio is 58:6058:60).).

Blackbody RadiationBlackbody Radiation

• One unsolved puzzle in physics in late nineteen One unsolved puzzle in physics in late nineteen

century was the spectral distribution of so calledcentury was the spectral distribution of so called

cavity radiationcavity radiation, also referred to as, also referred to as blackbody blackbody

radiationradiation. .

• It was shown by It was shown by KirchhoffKirchhoff that the most efficient that the most efficient

radiator of radiator of electromagnetic waveselectromagnetic waves was also a most was also a most

efficient absorber. A “perfect” absorber would be efficient absorber. A “perfect” absorber would be

one that absorbed all incident radiation.one that absorbed all incident radiation.

• Since no light would be reflected it is called aSince no light would be reflected it is called a

blackbodyblackbody. .

A small hole in the wall of the cavity approximating A small hole in the wall of the cavity approximating an ideal blackbody. Electromagnetic radiation (for an ideal blackbody. Electromagnetic radiation (for example, light) entering the hole has little chance of example, light) entering the hole has little chance of leaving before it is completely adsorbed within the leaving before it is completely adsorbed within the cavity.cavity.

Blackbody RadiationBlackbody Radiation• As the walls of the cavity absorb this incoming radiation , As the walls of the cavity absorb this incoming radiation ,

their temperature rises and begin to irradiate. their temperature rises and begin to irradiate.

• In 1879, the Austrian physicist In 1879, the Austrian physicist J.StefanJ.Stefan first measured the first measured the total amount of radiation emitted by blackbody at all total amount of radiation emitted by blackbody at all wavelengths and found it varied with absolute temperature. wavelengths and found it varied with absolute temperature.

• It was latter explained through a theoretical derivation by It was latter explained through a theoretical derivation by BoltzmanBoltzman, so the result became known as the , so the result became known as the Stefan-Stefan-Boltzman radiation Law:Boltzman radiation Law:

R = R = σσTT44

wherewhere RR is the power radiated is the power radiated per unit timeper unit time and and per unit per unit area of blackbodyarea of blackbody;; TT is the Kelvin temperature; and is the Kelvin temperature; and σσ is the is the Stefan-Boltzman constantStefan-Boltzman constant,, σσ=5.672 x 10=5.672 x 10-8 -8 W/mW/m22KK44..

Blackbody RadiationBlackbody Radiation

R = R = σσTT44 • Note that the power per unit area radiated by blackbody Note that the power per unit area radiated by blackbody

depends only on the temperature, and not of other depends only on the temperature, and not of other characteristic of the object, such as its color or the material, of characteristic of the object, such as its color or the material, of which it is composed. which it is composed.

• RR tells as the rate at which energy is emitted by the object. For tells as the rate at which energy is emitted by the object. For example, doubling the absolute temperature of an object example, doubling the absolute temperature of an object increases the energy flows out of the object by factor ofincreases the energy flows out of the object by factor of 2244=16=16..

• An object at room temperatureAn object at room temperature (300 K(300 K)) will double the rate at will double the rate at which it radiates energy as a result of temperature increase of which it radiates energy as a result of temperature increase of onlyonly 575700. .

Radiation emitted by the object at temperature Radiation emitted by the object at temperature TT that passed that passed through the slit is dispersed according to its wavelength. The through the slit is dispersed according to its wavelength. The prism shown would be an appropriate device for that part of prism shown would be an appropriate device for that part of the emitted radiation in the visible region. In other spectral the emitted radiation in the visible region. In other spectral regions other types of devices or wavelength-sensitive regions other types of devices or wavelength-sensitive detectors would be used. detectors would be used.

Spectral distribution functionSpectral distribution function R(R(λλ)) measured at different measured at different temperatures. Thetemperatures. The R(R(λλ)) axis is in arbitrary units for comparison axis is in arbitrary units for comparison only. Notice the range inonly. Notice the range in λλ of the visible spectrum. The Sun of the visible spectrum. The Sun emits radiation very close to that of a blackbody atemits radiation very close to that of a blackbody at 5800 K5800 K. . λλmm is indicated for theis indicated for the 5000-K5000-K and and 6000-K6000-K curves.curves.

Blackbody RadiationBlackbody Radiation

The German physicist The German physicist W.Wien W.Wien derived a relationship for derived a relationship for

maximum wavelength and absolute temperature, known as maximum wavelength and absolute temperature, known as

Wien’s displacement lawWien’s displacement law::

λλmmT = constant=(2.898 x 10T = constant=(2.898 x 10-3-3m)Km)K

ExampleExample: The wavelength at the peak of the spectral distribution : The wavelength at the peak of the spectral distribution

for a blackbody atfor a blackbody at 4300 K4300 K is is 674 nm674 nm (red). At what temperature (red). At what temperature

would the peak bewould the peak be 420 nm420 nm (violet)?(violet)?

Solution:Solution: From the Wien’s law, we have From the Wien’s law, we have

λλ11TT11 = = λλ22TT22

(674 x 10(674 x 10-9-9m)(4300 K) = (420 x 10m)(4300 K) = (420 x 10-9-9m)(Tm)(T22))

TT22=6900 K=6900 K

• This law is used to determine the surface This law is used to determine the surface temperatures of stars by analyzing their temperatures of stars by analyzing their radiation. It can also be used to map out the radiation. It can also be used to map out the variation in temperature over different regions of variation in temperature over different regions of the surface of an object. Such a map is called the surface of an object. Such a map is called thermographthermograph. .

• For example For example thermographthermograph can be used to detect can be used to detect cancer because cancerous tissue results in cancer because cancerous tissue results in increased circulation which produce a slight increased circulation which produce a slight increase in skin temperature. increase in skin temperature.

Radiation From the SunRadiation From the Sun.. The radiation emitted by The radiation emitted by the surface of the sun emits maximum power at the surface of the sun emits maximum power at wavelength of aboutwavelength of about 500 nm500 nm. . Assuming the sun to Assuming the sun to be a blackbody emitter, (a) what is it surface be a blackbody emitter, (a) what is it surface temperature? (b) Calculatetemperature? (b) Calculate λλmaxmax for a blackbody for a blackbody

at room temperature,at room temperature, T=300 KT=300 K..

The calculation of the The calculation of the distribution functiondistribution function R(R(λλ)) involves the calculation of involves the calculation of the energy density of the energy density of electromagnetic waves in electromagnetic waves in the cavity. The power the cavity. The power radiated out of the hole is radiated out of the hole is proportional to the total proportional to the total energy densityenergy density UU (energy (energy per unit volume) of the per unit volume) of the radiation in the cavity. Theradiation in the cavity. The

proportionality constant can be shown to beproportionality constant can be shown to be c/4c/4,, wherewhere cc is is the speed of the light:the speed of the light:

R = (1/4)cUR = (1/4)cU

Similarly, the spectral distribution of the power proportional to Similarly, the spectral distribution of the power proportional to the spectral distribution of the energy density in the cavity. the spectral distribution of the energy density in the cavity.

• IfIf u(u(λλ)d)dλλ is the fraction of is the fraction of the energy per unit volume the energy per unit volume in the cavity in the rangein the cavity in the range ddλλ, then, then u(u(λλ)) and and R(R(λλ)) are are related byrelated by

R(R(λλ)=(1/4)cu()=(1/4)cu(λλ)) The energy density The energy density

distribution functiondistribution function u(u(λλ)) can be calculated from can be calculated from classical physics. We can classical physics. We can find the number of modes find the number of modes of oscillation of the of oscillation of the electromagnetic field inelectromagnetic field in

the cavity with wavelength the cavity with wavelength λλ in the interval in the interval ddλλ and multiply it and multiply it by average energy per mode. The result is that the number by average energy per mode. The result is that the number of modes of oscillation per unit volume,of modes of oscillation per unit volume, n(n(λλ)),, is independent is independent of the shape of cavity and is given by:of the shape of cavity and is given by:

n(n(λλ) = 8) = 8ππccλλ-4-4

Rayleigh-Jeans EquationRayleigh-Jeans Equation

The number of modes of oscillation per unit volume:The number of modes of oscillation per unit volume:

n(n(λλ) = 8) = 8ππccλλ-4-4

According to the classical kinetic theory, the average energy According to the classical kinetic theory, the average energy

per mode of oscillation is per mode of oscillation is kTkT, the same as for a one-, the same as for a one-

dimensional harmonic oscillator, where dimensional harmonic oscillator, where kk is the is the Boltzman Boltzman

constantconstant. .

Classical theory thus predicts for the energy density spectral Classical theory thus predicts for the energy density spectral

distribution functiondistribution function

u(u(λλ) = kTn() = kTn(λλ)) = = 88ππckTckTλλ-4-4

Rayleigh-Jeans EquationRayleigh-Jeans Equation

u(u(λλ) = kTn() = kTn(λλ)) = = 88ππckTckTλλ-4-4

This prediction, initially derived by This prediction, initially derived by RayleighRayleigh, is called the , is called the Rayleigh-Jeans LawRayleigh-Jeans Law. .

At very long wavelength the At very long wavelength the Rayleigh-Jeans lawRayleigh-Jeans law agrees with agrees with experimentally determined spectral distribution, but at short experimentally determined spectral distribution, but at short wavelength this law predicts thatwavelength this law predicts that u(u(λλ)) becomes large, becomes large, approaching infinity asapproaching infinity as λ→λ→00, , whereas experiment shows that whereas experiment shows that the distribution actually approaches zero asthe distribution actually approaches zero as λ→λ→00. .

This enormous disagreement between experimental This enormous disagreement between experimental measurements and classical theory for short wavelength was measurements and classical theory for short wavelength was called thecalled the ultraviolet catastropheultraviolet catastrophe..

Comparison of the Rayleigh-Jeans Law with experimental Comparison of the Rayleigh-Jeans Law with experimental data at data at T=1600 KT=1600 K. The . The u(u(λλ)) axis is linear.axis is linear.

In 1900 the German In 1900 the German physicist physicist Max PlankMax Plank by by making some unusual making some unusual assumptions derived a assumptions derived a functionfunction u(u(λλ)) that agreed that agreed with experimental data.with experimental data. Classically, the Classically, the electromagnetic waves in electromagnetic waves in the cavity are produced the cavity are produced by accelerated electric by accelerated electric charges in the wallscharges in the walls

vibrating like simple harmonic oscillators. The average energy for vibrating like simple harmonic oscillators. The average energy for simple harmonic oscillator is calculated classically from simple harmonic oscillator is calculated classically from Maxwell-Maxwell-BoltzmanBoltzman distribution function: distribution function:

f(E) = Aef(E) = Ae-E/kT-E/kT

wherewhere AA is a constant andis a constant and f(E)f(E) is the fraction of oscillators with is the fraction of oscillators with energyenergy EE. .

Maxwell-Boltzman Maxwell-Boltzman distribution functiondistribution function

f(E) = Aef(E) = Ae-E/kT-E/kT

The average energy is then found, as is any weighted The average energy is then found, as is any weighted average:average:

Plank found that he could derive his empirical function Plank found that he could derive his empirical function

assuming the energy of oscillators, and hence the assuming the energy of oscillators, and hence the

radiation that they emitted, was a discrete variableradiation that they emitted, was a discrete variable that that

could take only the valuescould take only the values o, o, εε, 2, 2εε, 3, 3εε,…..,n,…..,nεε wherewhere nn is is

an integer, and that an integer, and that εε is proportional to the frequency of is proportional to the frequency of

the oscillators, and, thus, the radiation. the oscillators, and, thus, the radiation.

dEEAedEEEfE kT

E

Plank therefore wrote the energy asPlank therefore wrote the energy as

EEnn=n=nεε=nhf=nhf n=0,1,2,…..n=0,1,2,…..

wherewhere hh is a constant now calledis a constant now called PlankPlank constant constant. . The The Maxwell-BoltzmanMaxwell-Boltzman distribution than becomes:distribution than becomes:

ffnn= = AeAe-E/kT -E/kT = Ae= Ae-n-nεε/kT/kT

wherewhere AA is determined by normalization condition is determined by normalization condition that the sum of all fractionsthat the sum of all fractions ffnn must be equal 1. must be equal 1.

The normalization conditionThe normalization condition

The average energy of oscillation is then given by discrete-sum The average energy of oscillation is then given by discrete-sum equivalent equivalent

To solve this equation let putTo solve this equation let put

wherewhere y=ey=e-x-x.

kT

E

nnn AeEfEE

1 kT

n

n eAf

1....)1(

....])()([32

320

yyyA

eeeeAeAf xxxnxn

kT

hf

kTx

This sum is a series expansion ofThis sum is a series expansion of (1-y)(1-y)-1-1, , soso

(1-y)(1-y)-1-1=1+y+y=1+y+y22+y+y33+…+… ,,

thenthen ΣΣffnn=A(1-y)=A(1-y)-1-1=1=1 givesgives A=1-yA=1-y andand

Note thatNote that

so we haveso we have

sincesince

000

nxkT

nhf

kT

E

n neAhfnhfeAAeEE

1)1(. yeButedx

dne nxnxnx

221 )1()1()1(

yydx

dyyy

dx

de

dx

dne nxnx

yeedx

d

dx

dy xx )(

Plank’s LawPlank’s LawMultiplying this sum byMultiplying this sum by hfhf and usingand using A=(1-y)A=(1-y),, the the average energy is:average energy is:

Multiplying the numerator and the denominator byMultiplying the numerator and the denominator by eexx and substituting forand substituting for xx, we obtain:, we obtain:

x

xnx

e

hfe

y

hfyyyyhfnehfAE

11

)1()1( 2

11 kT

hf

kT

hf

e

hc

e

hfE

Plank’s LawPlank’s LawMultiplying this result by the number of oscillators Multiplying this result by the number of oscillators per unit volume in the intervalper unit volume in the interval ddλλ given bygiven by

n(n(λλ)=8)=8ππccλλ-4-4 (the number of modes of oscillation (the number of modes of oscillation per unit volume) we obtain the energy distribution per unit volume) we obtain the energy distribution function for the radiation in cavity:function for the radiation in cavity:

This function is calledThis function is called Plank’s LawPlank’s Law. .

1

8)(

52

kT

hc

e

hcu

The value of Plank’s constantThe value of Plank’s constant hh can be can be determined by fitting the functiondetermined by fitting the function

to the experimental data, although the direct to the experimental data, although the direct measurement is better, but more difficult. The measurement is better, but more difficult. The presently accepted value of Plank’ constant is: presently accepted value of Plank’ constant is:

h = 6.626 x 10h = 6.626 x 10-34-34 J J··ss = 4.136 x 10= 4.136 x 10-15-15 eV eV··ss

1

8)(

52

kT

hc

e

hcu

Plank’s LawPlank’s Law

Comparison of Plank’s Law and the Rayleigh-Jeans Law Comparison of Plank’s Law and the Rayleigh-Jeans Law with experimental data at with experimental data at T=1600 KT=1600 K. The . The u(u(λλ)) axis is linear.axis is linear.

4

8)(

ckT

u

A dramatic example of an application of A dramatic example of an application of Planck’s lawPlanck’s law is the is the test of the predictions of the so-called test of the predictions of the so-called Big Bang theoryBig Bang theory of the of the formation and expansion of the universe. formation and expansion of the universe.

Current cosmological theory suggests that the universe Current cosmological theory suggests that the universe originated in an extremely high-temperature explosion, one originated in an extremely high-temperature explosion, one consequence of which is to fill the infant universe with consequence of which is to fill the infant universe with radiation that can be approximate with black body spectral radiation that can be approximate with black body spectral distribution.distribution.

In 1965, In 1965, Arno PenziasArno Penzias and and Robert WilsonRobert Wilson discovered discovered radiation of wavelengthradiation of wavelength 7.35 cm7.35 cm reaching the Earth with reaching the Earth with same intensity from all directions in space. It was recognized same intensity from all directions in space. It was recognized soon as a remnant of the Big Bang (soon as a remnant of the Big Bang (relict radiationrelict radiation). ).

Plank’s LawPlank’s Law

The energy density spectral distribution of the cosmic The energy density spectral distribution of the cosmic microwave background radiation. The solid line is Plank’s Law microwave background radiation. The solid line is Plank’s Law with with T=2.735 KT=2.735 K. The measurements were made by the . The measurements were made by the Cosmic Back Ground Exploder (COBE) satellite. Cosmic Back Ground Exploder (COBE) satellite.

Problem 1Problem 1. . Thermal Radiation from the Human BodyThermal Radiation from the Human Body. . The temperature of the skin is approximately The temperature of the skin is approximately 3535°C°C. What . What is the wavelength at which the peak occurs in the is the wavelength at which the peak occurs in the radiation emitted from the skin?radiation emitted from the skin?

Problem 2.Problem 2. The Quantized OscillatorThe Quantized Oscillator. A . A 2-kg2-kg mass is mass is attached to a massless spring of force constant attached to a massless spring of force constant k=25N/mk=25N/m. The spring is stretched . The spring is stretched 0.4m0.4m from its from its equilibrium position and released. (a) Find the total equilibrium position and released. (a) Find the total energy and frequency of oscillation according to classical energy and frequency of oscillation according to classical calculations. (b) Assume that the energy is quantized calculations. (b) Assume that the energy is quantized and find the quantum number, and find the quantum number, nn, for the system. (c) How , for the system. (c) How much energy would be carried away in one-quantum much energy would be carried away in one-quantum change? change?

Problem 3Problem 3.. The Energy of a “Yellow” PhotonThe Energy of a “Yellow” Photon. What is . What is the energy carried by a quantum of light whose the energy carried by a quantum of light whose frequency equals frequency equals 6 x 106 x 101414 Hz Hz yellow light? What is the yellow light? What is the wavelength of this light? wavelength of this light?

The Photoelectric EffectThe Photoelectric EffectIt is one of the ironies in the history of the science that It is one of the ironies in the history of the science that

in the same famous experiment of in the same famous experiment of Heinrich HertzHeinrich Hertz in1887 in in1887 in which he produced and detected electromagnetic waves, which he produced and detected electromagnetic waves, thus confirmed thus confirmed Maxwell’sMaxwell’s wave theory of light, he also wave theory of light, he also discovered the photoelectric effect led directly to particle discovered the photoelectric effect led directly to particle description of light.description of light.

It was found that negative charged particles were It was found that negative charged particles were emitted from a clean surface when exposed to lightemitted from a clean surface when exposed to light..

P.Lenard in 1900 detected them in a magnetic field P.Lenard in 1900 detected them in a magnetic field and found that they had a and found that they had a charge-to-mass ratiocharge-to-mass ratio of the same of the same magnitude as that measured by Thompson for cathode rays: magnitude as that measured by Thompson for cathode rays: the particles being emitted were electrons. the particles being emitted were electrons.

Photoelectrons going through the hole in anode Photoelectrons going through the hole in anode AA are recorded by the electrometer connected to are recorded by the electrometer connected to αα.. A A magnetic field, indicated by the circular pole piece, magnetic field, indicated by the circular pole piece, could deflect the particles to an electrometer could deflect the particles to an electrometer connected to connected to ββ, enabling the establishment of the , enabling the establishment of the sign of their charge and their sign of their charge and their q/mq/m ratio.ratio.

Schematic diagram Schematic diagram of the apparatus of the apparatus used by P.Lenard to used by P.Lenard to demonstrate the demonstrate the photoelectric effect photoelectric effect and to show that the and to show that the particles emitted in particles emitted in the process were the process were electrons. Light from electrons. Light from the source the source LL strikes strikes the cathode the cathode CC..

increased by making the anode positive with respect to increased by making the anode positive with respect to cathode. cathode.

LettingLetting VV be the potential difference betweenbe the potential difference between AA and and CC the the next picture shows the current versusnext picture shows the current versus VV for two values of for two values of the intensity of light incident on the cathode:the intensity of light incident on the cathode:

If some of emitted If some of emitted electrons that reaches electrons that reaches an anodean anode AA pass pass through the small hole, through the small hole, a current results in the a current results in the external electrometer external electrometer circuit connected tocircuit connected to αα..

The number of emitted The number of emitted electrons, reaching the electrons, reaching the anode, can beanode, can be

Photocurrent Photocurrent ii versus anode voltageversus anode voltage VV for light of frequencyfor light of frequency ff with two intensitieswith two intensities II11 andand II2 2 , , wherewhere II22>I>I11 . . At sufficiently large At sufficiently large

VV all emitted electrons reach the anode and the current all emitted electrons reach the anode and the current reaches its maximum value. reaches its maximum value. From experiment it was observed that the maximum current From experiment it was observed that the maximum current is proportional to the light intensity. is proportional to the light intensity.

An expected result – if the intensity of incident light doubled An expected result – if the intensity of incident light doubled the number of emitted electrons should also double. If the number of emitted electrons should also double. If intensity of incident light is too low to provide electrons with intensity of incident light is too low to provide electrons with energy necessary to escape from the metal, no emission of energy necessary to escape from the metal, no emission of electron should be observed. electron should be observed. However, there was no minimum intensity below which the However, there was no minimum intensity below which the current was absent. current was absent.

When When VV is negative, the electrons are repelled from the is negative, the electrons are repelled from the anode and only electrons with initial kinetic energyanode and only electrons with initial kinetic energy mvmv22/2/2 grater thangrater than eVeV can reach the anode. From the graph we can can reach the anode. From the graph we can see if the voltage is less, thansee if the voltage is less, than –V–V00 no electrons reach anode. no electrons reach anode.

The potentialThe potential VV00 is called theis called the stopping potentialstopping potential. . It related to It related to

the maximum kinetic energy asthe maximum kinetic energy as

(mv(mv22/2)/2)maxmax = eV = eV00

The experimental result, thatThe experimental result, that VV00 is independent of the is independent of the

incident light intensity was surprising – increasing the incident light intensity was surprising – increasing the rate of energy falling on the cathode does not increase rate of energy falling on the cathode does not increase the maximum kinetic energy of the emitted electronsthe maximum kinetic energy of the emitted electrons. .

In 1905, Einstein offered an explanation of this result: In 1905, Einstein offered an explanation of this result: he assumed, thathe assumed, that energy quantization used by Plank energy quantization used by Plank in the blackbody problem was a universal in the blackbody problem was a universal characteristic of lightcharacteristic of light.. Light energy consist of discrete Light energy consist of discrete

quanta of energyquanta of energy hfhf..

When one of this quanta, calledWhen one of this quanta, called photonphoton, penetrates the , penetrates the surface of cathode, all of its energy may be given surface of cathode, all of its energy may be given completely to electron. Ifcompletely to electron. If ΦΦ is the energy necessary to is the energy necessary to remove an electron from the surface (remove an electron from the surface (ΦΦ is called theis called the work work functionfunction and is a characteristic of the metal), the and is a characteristic of the metal), the maximum kinetic energy of the electrons leaving the maximum kinetic energy of the electrons leaving the surface will besurface will be (hf – (hf – ΦΦ)) and the stopping potentialand the stopping potential VV00 should should

be given bybe given by

This equation is referred as theThis equation is referred as the photoelectric effect photoelectric effect equation. equation.

hfmv

eV

max

2

0 2

As can be seen fromAs can be seen from

the slopethe slope of the line on the graph of the line on the graph VV00 versus versus ff should equalshould equal h/eh/e. .

The minimum, or threshold, frequency for photoelectric effect,The minimum, or threshold, frequency for photoelectric effect, fftt and the and the

corresponding threshold wavelengthcorresponding threshold wavelength λλtt are related to work functionare related to work function ΦΦ by settingby setting VV00

= 0= 0 : :

Photons of frequency lower thanPhotons of frequency lower than fftt ( and therefore having wavelength grater than( and therefore having wavelength grater than

λλtt ) ) do not have enough energy to eject an electron from the metal. do not have enough energy to eject an electron from the metal.

hfmv

eV

max

2

0 2

tthc

hf

For constant For constant II Einstein’s explanation of the photoelectric effect Einstein’s explanation of the photoelectric effect indicates that the magnitude of the stopping voltage should be indicates that the magnitude of the stopping voltage should be grater forgrater for ff22 thanthan ff11, as observed, and that there should be a , as observed, and that there should be a

threshold frequencythreshold frequency fftt below which no photoelectrons were below which no photoelectrons were

seen, also in agreement with experiment.seen, also in agreement with experiment.

Millikan’s data for stopppping potential versus frequency for the

photoelectric effect. The data falls on a straight line of slope

h/e, as predicted by Einstein. The intercept of the stopping

potential axis is –Ф/e.

Example: Photoelectric Effect in PotassiumExample: Photoelectric Effect in Potassium

The threshold wavelength of potassium is The threshold wavelength of potassium is 558 nm558 nm. What is . What is the work function for potassium? What is the stopping the work function for potassium? What is the stopping potential when light of wavelength potential when light of wavelength 400 nm400 nm is used? is used?

Example: Photoelectric Effect in PotassiumExample: Photoelectric Effect in Potassium

The threshold wavelength of potassium is The threshold wavelength of potassium is 558 nm558 nm. What is the . What is the work function for potassium? What is the stopping potential work function for potassium? What is the stopping potential when light of wavelength when light of wavelength 400 nm400 nm is used? is used?

• Solution: Solution:

eVnm

nmeVhc

e

hc

e

hf

e

ee

hfV

hfmveV

t

t

t

22.2558

1240

2

1

0

max

20

Example: Photoelectric Effect in PotassiumExample: Photoelectric Effect in Potassium

• The threshold wavelength of potassium is The threshold wavelength of potassium is 558 nm558 nm. . What is the work function for potassium? What is What is the work function for potassium? What is the stopping potential when light of wavelength the stopping potential when light of wavelength 400 nm400 nm is used? is used?

VeVeV

eVnm

nmeVhcV

88.022.210.3

22.2400

12400

Photoelectric Effect for SodiumPhotoelectric Effect for Sodium

A sodium surface is illuminated with light of A sodium surface is illuminated with light of wavelength wavelength 3x103x10-7-7mm. The work function for sodium . The work function for sodium is is 2.28 eV2.28 eV. Find (a) the kinetic energy of the . Find (a) the kinetic energy of the ejected photoelectrons and (b) the cutoff ejected photoelectrons and (b) the cutoff wavelength for sodium.wavelength for sodium.

X Rays and the Compton EffectX Rays and the Compton Effect

Future evidence of the correctness of the photon Future evidence of the correctness of the photon concept was given by concept was given by Arthur ComptonArthur Compton, who measured the , who measured the scattering ofscattering of x-raysx-rays by free electrons.by free electrons.

The German physicist The German physicist Wilhelm RWilhelm Rööentgenentgen discovered discovered x-raysx-rays in 1895 when he was working with a cathode-ray tube. in 1895 when he was working with a cathode-ray tube. He found, that “rays”, originating from the point where the He found, that “rays”, originating from the point where the cathode rays (electrons) hit the glass tube, or a target within the cathode rays (electrons) hit the glass tube, or a target within the tube, could pass through the materials opaque to light and tube, could pass through the materials opaque to light and activate a fluorescent screen or photographic film. He found activate a fluorescent screen or photographic film. He found that all materials were transparent to this rays to some degree, that all materials were transparent to this rays to some degree, depending of the density of this materials.depending of the density of this materials.

The slight diffraction of an The slight diffraction of an x-rayx-ray beam after passing slit beam after passing slit of a few thousands of a mm wide indicated their wavelength in of a few thousands of a mm wide indicated their wavelength in other ofother of 1010-10-10m = 0.1nmm = 0.1nm. .

(a) Early (a) Early x-rayx-ray tube and (b) typical of the mid-twenties tube and (b) typical of the mid-twenties century century x-rayx-ray tube design. tube design.

(a) (b)

Diagram of the components of a modernDiagram of the components of a modern x-ray x-ray tube. Design tube. Design technology has advanced enormously, enabling very high technology has advanced enormously, enabling very high operating voltages, beam currents, and operating voltages, beam currents, and x-rayx-ray intensities, but intensities, but the essential elements of the tubes remain unchanged.the essential elements of the tubes remain unchanged.

Experiment soon confirmed that Experiment soon confirmed that x-raysx-rays are a form are a form of electromagnetic radiation with wavelength of of electromagnetic radiation with wavelength of aboutabout 0.01nm 0.01nm toto 0.10 nm 0.10 nm..

It was also known that atoms in crystals are It was also known that atoms in crystals are arranged in regular arrays that are spaced by about arranged in regular arrays that are spaced by about same distances.same distances.

In 1912, Laue suggested that since the wavelength In 1912, Laue suggested that since the wavelength ofof x-raysx-rays were on the same order of magnitude as were on the same order of magnitude as the spacing of atoms in a crystal, the regular array the spacing of atoms in a crystal, the regular array of atoms in crystal might act as a three-dimensional of atoms in crystal might act as a three-dimensional grating for diffraction ofgrating for diffraction of x-raysx-rays. .

X-Ray’s DiffractionX-Ray’s Diffraction

• W.L.BraggW.L.Bragg, in 1912, proposed a simple and and , in 1912, proposed a simple and and convenient way of analyzing the diffraction of convenient way of analyzing the diffraction of x-x-raysrays due to scattering from various sets of due to scattering from various sets of parallel planes of atoms, now called parallel planes of atoms, now called Bragg Bragg planes.planes.

• Two sets of Two sets of Bragg planesBragg planes are illustrated for are illustrated for NaClNaCl,, which has a simple crystal structure called which has a simple crystal structure called face-centered cubicface-centered cubic..

X-Ray’s DiffractionX-Ray’s Diffraction

A face-centered cubicA face-centered cubic crystal of crystal of NaClNaCl showing showing two sets of two sets of Bragg planesBragg planes..

• Waves scattered at equal angels from atoms in Waves scattered at equal angels from atoms in two different planes will be in phase (constructive two different planes will be in phase (constructive interference) if the difference in path length is an interference) if the difference in path length is an integer number of wavelength. This condition is integer number of wavelength. This condition is satisfied ifsatisfied if

2d sin 2d sin θθ = m = mλλ where where mm = = an integeran integer

This equation called theThis equation called the Bragg conditionBragg condition..

Bragg scattering from two successive planes. The waves from Bragg scattering from two successive planes. The waves from the two atoms shown have a path difference ofthe two atoms shown have a path difference of 2dSin2dSinθθ. They . They

will be in phase if the Bragg conditionwill be in phase if the Bragg condition 2dSin2dSinθθ = m = mλλ is met.is met.

Schematic sketch of the Schematic sketch of the LaueLaue experiment. The crystal acts as a experiment. The crystal acts as a three-dimensional grating, which diffracts thethree-dimensional grating, which diffracts the x-rayx-ray beam and beam and produce a regular array of spots, called a produce a regular array of spots, called a Laue patternLaue pattern, on a , on a photographic plate.photographic plate.

Modern Modern LaueLaue-type-type x-ray x-ray diffraction patterndiffraction pattern using a using a niobium diboride crystal niobium diboride crystal andand 20-kV20-kV molybdenum molybdenum x-raysx-rays. [General Electric . [General Electric Company.]Company.]

Incident Incident X-rayX-rayBeamBeam

Scattered

X-rays

Schematic diagram of Schematic diagram of Bragg crystal spectrometerBragg crystal spectrometer. A collimated . A collimated x-rayx-ray beam is incident on a crystal and scattered into an beam is incident on a crystal and scattered into an ionization chamber. The crystal and ionization chamber can be ionization chamber. The crystal and ionization chamber can be rotated to keep the angles of incidence and scattering equal as rotated to keep the angles of incidence and scattering equal as both are varied. By measuring the ionization in the chamber as a both are varied. By measuring the ionization in the chamber as a function of angle, the spectrum of the function of angle, the spectrum of the x-raysx-rays can be determined can be determined using the Bragg condition using the Bragg condition 2dSin2dSinθθ = m = mλλ, where , where dd is the is the separation of the Bragg planes in the crystal. If the wavelength separation of the Bragg planes in the crystal. If the wavelength λλ is known, the spacing is known, the spacing dd can be determined.can be determined.

Two typical Two typical x-rayx-ray spectra are produced by accelerating spectra are produced by accelerating electrons through two voltageselectrons through two voltages VV and bombarding a tungsten and bombarding a tungsten target.target. I(I(λλ)) is the intensity emitted with the wavelength intervalis the intensity emitted with the wavelength interval ddλλ at each value ofat each value of λλ..

The spectrum consist of a series of sharp lines, called theThe spectrum consist of a series of sharp lines, called the characteristic spectrumcharacteristic spectrum. The line spectrum is a . The line spectrum is a characteristic of target material and varies from element to characteristic of target material and varies from element to element.element.

The continuous spectrum has a sharp cutoff wavelengthThe continuous spectrum has a sharp cutoff wavelength λλmm

which is independent of the target material but depends on which is independent of the target material but depends on the energy of the bombarding electrons.the energy of the bombarding electrons.

If the voltage of theIf the voltage of the x-rayx-ray tube istube is VV in volts, the cutoff in volts, the cutoff wavelength was found empirically bywavelength was found empirically by

It was pointed out by EinsteinIt was pointed out by Einstein thatthat x-rayx-ray production by production by electron bombardment was an inverse photoelectric effect and electron bombardment was an inverse photoelectric effect and equationequation

should be applied. Theshould be applied. The λλmm simply correspond to a photon with simply correspond to a photon with

the maximum energy of the electrons, i.e. the photon emitted the maximum energy of the electrons, i.e. the photon emitted when the electron losses all of its kinetic energy in a simple when the electron losses all of its kinetic energy in a simple collision.collision.

nmVm

31024.1

hfmv

eV

max

2

0 2

• Since the kinetic energy of the electron in theSince the kinetic energy of the electron in the x-rayx-ray tube istube is 20,000 eV20,000 eV or larger, the or larger, the work functionwork function ΦΦ is negligible by comparison and equation is negligible by comparison and equation

becomesbecomes

oror

Thus, theThus, the x-rayx-ray spectrum can be explained by Plank’s quantum hypothesis andspectrum can be explained by Plank’s quantum hypothesis and λλmm can be used to determinecan be used to determine h/eh/e. .

hfmv

eV

max

2

0 2

nmeV

nmeV

eV

hc 31024.11240

hc

hfeV

Schematic sketch of Compton apparatus.Schematic sketch of Compton apparatus. X-raysX-rays from the tube from the tube strike the carbon blokestrike the carbon bloke RR and are scattered into a and are scattered into a BraggBragg-type -type crystal spectrometer. In this diagram, the scattering angle iscrystal spectrometer. In this diagram, the scattering angle is 303000. . The beam was defined by slitsThe beam was defined by slits SS11 andand SS22. . Although the entire Although the entire

spectrum is being scattered byspectrum is being scattered by RR, the spectrometer scanned the , the spectrometer scanned the region around region around KKαα line of molybdenum.line of molybdenum.

Compton EffectCompton Effect

Derivation of Compton’s EquationDerivation of Compton’s Equation LetLet λλ11 andand λλ22 be the wavelengths of the incident and scatteredbe the wavelengths of the incident and scattered

x raysx rays, respectively. The corresponding momentum are , respectively. The corresponding momentum are

andand

using using ffλλ = c = c. . Since Since ComptonCompton used the used the KKαα line of molybdenumline of molybdenum

(( λλ = = 0.0711 nm0.0711 nm), the energy of the incident), the energy of the incident x rayx ray ( (17.4 keV17.4 keV) is ) is much greater than the binding energy of the valence electrons in much greater than the binding energy of the valence electrons in the carbon scattering block (aboutthe carbon scattering block (about 11 eV11 eV); therefore, the carbon ); therefore, the carbon electron can be considered to be free. electron can be considered to be free.

1

111

h

c

hf

c

EP

2

22

h

c

EP

Conservation of momentum givesConservation of momentum gives

wherewhere ppee is the momentum of the electron after the is the momentum of the electron after the

collision andcollision and θθ is the scattering angle for the photon,is the scattering angle for the photon, measured as shown in Figure.measured as shown in Figure. The energy of the electron The energy of the electron

before the collision is simply its rest energybefore the collision is simply its rest energy EE0 0 = mc= mc22..

cos22 2122

2122

21

2

21

21ppppppppp

ppp

e

e

Diagram for Derivation of Compton’s EquationDiagram for Derivation of Compton’s Equation

After the collision, the energy of the electron isAfter the collision, the energy of the electron is (E(E0022 + p + pee

22cc22))1/21/2..

Conservation of energy givesConservation of energy gives

pp11c + Ec + E00 = p = p22c + (Ec + (E002 2 + p+ pee

22cc22))1/21/2

Transposing the termTransposing the term pp22cc and squaring we obtainand squaring we obtain

EE0022+c+c22(p(p11-p-p22))22+2cE+2cE00(p(p11-p-p22)=E)=E00

22+p+pee22cc22

oror

ppee22=p=p11

22+p+p2222-2p-2p11pp22+2E+2Eoo(p(p11-p-p22)/c)/c

ppee22=p=p11

22+p+p2222-2p-2p11pp22+2E+2Eoo(p(p11-p-p22)/c)/c

If we eliminateIf we eliminate ppee22 from the previous equations, we obtainfrom the previous equations, we obtain

EE00(p(p11-p-p22)/c = p)/c = p11pp22(1-cos(1-cosθθ))

Multiplying each term byMultiplying each term by hc/phc/p11pp22EE00 and usingand using λλ=h/p=h/p, , we we

obtainobtain Compton’s equationCompton’s equation::

λλ22 – – λλ11 = hc(1-cos = hc(1-cosθθ)/E)/E00 = hc(1-cos = hc(1-cos θθ)/mc)/mc22

oror

λλ2 2 – – λλ11 = h(1-cos = h(1-cos θθ)/mc)/mc

Compton WavelengthCompton Wavelength

)cos1(12 cm

hc

e

The increase in wavelengths is independent of the The increase in wavelengths is independent of the wavelengthwavelength λλ11 of the incident photon. The quantity of the incident photon. The quantity

has dimensions of length and is called thehas dimensions of length and is called the

Compton WavelengthCompton Wavelength. Its value is. Its value is cm

h

e

pmmeV

nmeV

cm

hc

cm

h

eeC 43.21043.2

1011.5

1240 1252

Compton WavelengthCompton Wavelength

BecauseBecause λλ22–λ–λ11 is small it is difficult to observeis small it is difficult to observe

unlessunless λλ11 is so small that the fractional changeis so small that the fractional change

(λ(λ22–λ–λ11)/λ)/λ11 is appreciable. Compton usedis appreciable. Compton used X-raysX-rays of of

wavelengthwavelength 71.1pm71.1pm. The energy of a . The energy of a photonphoton of this of this wavelength is wavelength is

keVnm

nmeVhcE 1.17

0711.0

1240

Since this is mush greater than the binding energy of Since this is mush greater than the binding energy of the valence electrons in most atoms, these electrons the valence electrons in most atoms, these electrons can be considered to be essentially free. Compton’s can be considered to be essentially free. Compton’s measurements confirmed the correctness of the measurements confirmed the correctness of the photon concept. photon concept.

Reflection from CalciteReflection from Calcite

If the spacing between certain planes in crystal If the spacing between certain planes in crystal of calcite isof calcite is 0.314nm0.314nm, find the grazing angles at , find the grazing angles at which first and third order interference will occur which first and third order interference will occur forfor x-raysx-rays of wavelengthof wavelength 0.070nm0.070nm..

Compton Scattering at 45Compton Scattering at 4500

X-raysX-rays wavelengthwavelength λλ00=0.200 000 nm=0.200 000 nm are are

scattered from a block of material. The scatteredscattered from a block of material. The scattered x-raysx-rays are observed at an angle ofare observed at an angle of 454500 to the to the incident beam. Calculate the wavelength of theincident beam. Calculate the wavelength of the x-x-raysrays scattered at this angle. scattered at this angle.

The Minimum X-ray WavelengthThe Minimum X-ray Wavelength

Calculate the minimum x-rays wavelength Calculate the minimum x-rays wavelength produced when electrons are accelerated through a produced when electrons are accelerated through a potential difference ofpotential difference of 100 000V100 000V, a not-uncommon , a not-uncommon voltage for anvoltage for an x-rayx-ray tube.tube.

Photoelectric Effect in LithiumPhotoelectric Effect in Lithium

Light of wavelength ofLight of wavelength of 400nm400nm is incident upon lithium is incident upon lithium ((Φ = 2.9eVΦ = 2.9eV). Calculate (a) the photon energy and (b) ). Calculate (a) the photon energy and (b) the stopping potentialthe stopping potential VV00..

What frequency of light is needed to produce electrons What frequency of light is needed to produce electrons of kinetic energyof kinetic energy 3eV3eV from illumination fromfrom illumination from LiLi??