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Physics 5110 HW #04 Solutions 1. Note some students used A=179 because of my typographic error. The equivalent plots are given in appendix B. They should not be penalized for this.
(a) a = 1.21 A1/3 fm = 1.21 fm (197)1/3=1.21 fm (5.818647867) = 7.04 fm
ρ0=3Ze/4πa3=3(79)(1.6×10-19C)/[4(3.1415926)(7.04 fm)3]=8.65×10-21 C/fm3.
NOTE in the original posting I mistakenly put A=179…those who used this value were told not to re-do the problem (it’s just a number! And Au-179 can be made!) For A=179:
a = 1.21 A1/3 fm = 1.21 fm (179)1/3=1.21 fm (5.635740795) = 6.82 fm
ρ0=3Ze/4πa3=3(79)(1.6×10-19C)/[4(3.1415926)(6.82 fm)3]=9.52×10-21 C/fm3.
Both are acceptable for full credit.
(b) From the textbook Eqn 2.18: drrqrrqZe
qF
= ∫
∞
h
h
0
2 sin)(4)( ρπ
We will do the integrals numerically using EXCEL, where we first define the integrand
( )
−
+
=d
ar
e
drrqrqrIntegrand1
)/(sin)/;( hh , and the integral: drqrIntegradqIntegral ∫
∞
=0
)/;()/( hh
and the integral is approximated using the simplest method: Midpoint Rule Integration:
rqrIntegradqIntegralN
nn ∆≈ ∑
=
)/;()/(1
hh , where rnrn ∆
−=
21
I wanted people to use at least 1000 steps and at least up to a+5d. Here a+5d≈11 fm (for either value of A). So I chose ∆r = 0.01 fm, I my case I chose N=2000, but N≈1100 would be OK.
One important step is to calculate q in convenient units, with q taken as the square-root of the q2 values you were told to use: 0.10→0.99(GeV/c)2 in steps of 0.01 (GeV/c)2. Clearly since we are using units of fm for r we want in fm-1 because the arguments to sine should be dimensionless. Now =
6.57×10-16 eV-s. The calculation for
h/
h/q h2
2
=
cGeVηq is given by:
fmseVsfmc
GeVeV
cGeV
seVcGeVq 107.51
1057.61
/1000.3101
1057.61
1623
9
16 ηηη ×=⋅×
××
××=⋅×
×= −−h
In my case, this formula used in the calculation is put into the cells for the values of q.
Note that in my definition of the integral above I left out ρ0, which must now be included when calculating the form factor itself: )( 2qF
3
1
302 )/(3)/(
3/44)/(4)(
aqIntegralqqIntegral
aZe
qZeqIntegral
qZeqF h
hh
hh
h−
=××
=××
=
ππρπ
h/q)( 2qF
Note
about units. As defined the integral has units of fm2, fm-1, and a3 fm3. This makes the form factor dimensionless, as it SHOULD be!
The result for : tabulated and plotted )( 2qF
q^2 (GeV/c)^2 F(q^2)
q^2 (GeV/c)^2 F(q^2)
q^2 (GeV/c)^2 F(q^2)
q^2 (GeV/c)^2 F(q^2)
0.1 -2.50E-03 0.33 4.41E-05 0.56 3.55E-06 0.79 -1.28E-06 0.11 -3.13E-03 0.34 6.38E-05 0.57 5.25E-06 0.8 -9.82E-07 0.12 -2.70E-03 0.35 7.31E-05 0.58 6.36E-06 0.81 -6.86E-07 0.13 -1.78E-03 0.36 7.37E-05 0.59 6.92E-06 0.82 -4.02E-07 0.14 -7.84E-04 0.37 6.76E-05 0.6 7.00E-06 0.83 -1.42E-07 0.15 3.06E-05 0.38 5.69E-05 0.61 6.69E-06 0.84 8.80E-08 0.16 5.67E-04 0.39 4.36E-05 0.62 6.06E-06 0.85 2.83E-07 0.17 8.24E-04 0.4 2.94E-05 0.63 5.21E-06 0.86 4.40E-07 0.18 8.53E-04 0.41 1.57E-05 0.64 4.23E-06 0.87 5.59E-07 0.19 7.30E-04 0.42 3.54E-06 0.65 3.18E-06 0.88 6.40E-07
0.2 5.25E-04 0.43 -6.53E-06 0.66 2.13E-06 0.89 6.87E-07 0.21 2.99E-04 0.44 -1.41E-05 0.67 1.15E-06 0.9 7.01E-07 0.22 9.28E-05 0.45 -1.92E-05 0.68 2.54E-07 0.91 6.87E-07 0.23 -6.97E-05 0.46 -2.20E-05 0.69 -5.13E-07 0.92 6.50E-07 0.24 -1.79E-04 0.47 -2.26E-05 0.7 -1.14E-06 0.93 5.95E-07 0.25 -2.36E-04 0.48 -2.16E-05 0.71 -1.61E-06 0.94 5.25E-07 0.26 -2.49E-04 0.49 -1.93E-05 0.72 -1.94E-06 0.95 4.45E-07 0.27 -2.29E-04 0.5 -1.61E-05 0.73 -2.13E-06 0.96 3.60E-07 0.28 -1.87E-04 0.51 -1.24E-05 0.74 -2.20E-06 0.97 2.74E-07 0.29 -1.34E-04 0.52 -8.63E-06 0.75 -2.15E-06 0.98 1.89E-07
0.3 -7.87E-05 0.53 -4.95E-06 0.76 -2.02E-06 0.99 1.08E-07 0.31 -2.84E-05 0.54 -1.61E-06 0.77 -1.82E-06 0.32 1.32E-05 0.55 1.25E-06 0.78 -1.56E-06
F(q^2)
-3.50E-03
-3.00E-03
-2.50E-03
-2.00E-03
-1.50E-03
-1.00E-03
-5.00E-04
0.00E+00
5.00E-04
1.00E-03
1.50E-03
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
F(q^2)
The two following plots are NOT required: the first is with the vertical scale zoomed in. The
second is a plot of
)( 2qF22 )(qF which shows up the oscillatory structure better
F(q^2)
-1.00E-04
-8.00E-05
-6.00E-05
-4.00E-05
-2.00E-05
0.00E+00
2.00E-05
4.00E-05
6.00E-05
8.00E-05
1.00E-04
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
F(q^2)
|F(q^2)|^2
1.00E-15
1.00E-14
1.00E-13
1.00E-12
1.00E-11
1.00E-10
1.00E-09
1.00E-08
1.00E-07
1.00E-06
1.00E-05
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
|F(q^2)|^2
(c) The electron-gold scattering differential cross section is give by:
( )2279
MottAuqF
Z
dd
edd
×=
Ω=
→
Ωσσ ,
where ( )2qF was numerically calculated in part (b)
From your textbook the inclusion of spin ½ for the electron gives an additional factor to the Rutherford scattering:
[ ] )2/(cos4Rutherford
)2/(sin14Rutherford1/2spin
222 θσθβσσ
Ω≈−
Ω=
Ω dd
dd
dd
where we have made the (very good) approximation that β = v/c ≈ 1 for 0.5 GeV electrons. Note that your textbook does not carry the factor of 4 in the equation, because it cancels exactly the factor of 4 that you would have to divide by in changing the Rutherford formula given in the book (which was for explicitly for alpha particles), and the Mott scattering (and spin ½) formula was explicitly for electrons. But if you just combined the two equations in the book together you get the same answer as I have here. Note also I have neglected the recoil in calculating Mott scattering, because the factor q2/M2c2 ≤ 3×10−5. We used:
( ) )2/(cos)2/(sin4Mott
242
2222
θθ
ασE
cZzZz
dd h
≈→
Ω, z = −1, Z = 79.
Not required: Mott Scattering dσ/dΩ for target Z=79 and electron beam energy E=0.5 GeV
The differential cross section calculated for electron scattering on gold vs. scat. angle in degrees.
Mott C.S. (fm^2)
1.00E-04
1.00E-03
1.00E-02
1.00E-01
1.00E+00
1.00E+01
1.00E+02
1.00E+03
0.00 30.00 60.00 90.00 120.00 150.00 180.00
Mott C.S. (fm^2)
And lastly the actual differential cross section obtained by multiplying the Mott cross-section above by
ds/dW (fm^2)
1.00E-19
1.00E-16
1.00E-13
1.00E-10
1.00E-07
1.00E-04
0.00 30.00 60.00 90.00 120.00 150.00 180.00
ds/dW (fm^2)
Problem 2: (a) As defined, ρ0 should have units of C/fm5 (mea culpa) given the definition:
)/exp()()( 20 arCrr −−= ρρ
The neutron has zero charge:
( ) [ 0)2()4(40
4 )( 02 =−=
∞= ∫ CAAdrrrnnQ πρπρ ] , where ( ) 1 !
0exp )( +⋅=
∞≡ ∫ kakdr-r/akrkA
for k ≥ 0 (easily demonstrated by integration by parts), and so:
[ ] [ ] [ ] 0128!2!44)2()4(4 530
3500 =−=−=−= CaaCaaCAAnQ πρπρπρ
)/exp()12()( 22 ararr −−= ρρ212aC =
(b) We are given from o
(∫∞
≡0
4 )(1 22 rrrnenr πρ
Just applying the definiti
( )22
0 4 )(1 drrrrne
πρ =∞
∫
702 1728a
enrπρ
=
Note that this means 0ρ for the charge Qn above w
(c) The form factor is the
∞= ∫
0)( 4)( 2 rnr
eqqnF h ρπ
This integral can be evalu> assume(a>0); > assume(Omega>0); > int(r*(r^2-12*a^2)*
−24 a~7 Ω∼3 ( + 3 Ω∼2 a
( ) + 1 Ω∼2 a~2 4
The “~” denotes the a varia
0
ther measurements:
) −= fm 116.0 22 dr
on but with Z=1 normalization:
[ ] [ ] [ ] 70707020 1728288720
4!412!6
4)4(12)6(
4a
ea
ea
eAaA
eπρπρπρπρ
=−=⋅−=−
/1C 10419.7/1
1728fm 116. 2242
×−
= −
ρπρa fm30
⋅
⋅−
=e 7
00
is a negative quantity! The value of a has to be positive otherwise the integral ould have diverged.
n given by:
∞
−−
=
∫
−
0sin)/exp()12(
4sin 22
10 drqrararrq
edrqr
hhh
πρ
ated using MAPLE, where we have denoted h/q≡Ω
exp(-r/a)*sin(Omega*r),r=0..infinity);
)~2
ble with “assume”ed properties. Or we can do this the Macho way (see appendix A)
Moving on:
[ ][ ]
[ ][ ]42
2250
42
234
102
)/(1)/(3)/(96
)/(1)/(3)/(244)(
h
hh
h
hh
h qaqaqa
ea
qaqaqaaq
eqnF
+
+−=
+
+−
=
− πρπρ
where 7/1fmC 10419.3
0
224
⋅×−=
−
ρa , and
I started with ρ0 = −1×10−19 C/fm5 (this is roughly the elementary charge over a length scale of 1 fm), which gave values of which were too small. Next I tried ρ0 = −1×10−18 C/fm5, and this gave
values which were too large. Values in the range −(3→5)×10−19 C/fm5 run through the data points mostly within the error bars. I chose −4×10−19 C/fm5 because it gave roughly the same number of points above and below the fit line. Anything close is acceptable. (See attached EXCEL spreadsheet)
)( 2qnF
Table of Fit values (not required): )( 2qnF
q^2 (GeV/c)^2 Fit F(q^2)
q^2 (GeV/c)^2 Fit F(q^2)
q^2 (GeV/c)^2 Fit F(q^2)
q^2 (GeV/c)^2 Fit F(q^2)
0.00 0.000000 0.52 0.063014 1.04 0.046783 1.56 0.0328230.02 0.009312 0.54 0.062613 1.06 0.046146 1.58 0.0323930.04 0.017445 0.56 0.062165 1.08 0.045516 1.60 0.0319690.06 0.024542 0.58 0.061677 1.10 0.044894 1.62 0.0315530.08 0.030725 0.60 0.061152 1.12 0.044280 1.64 0.0311430.10 0.036103 0.62 0.060597 1.14 0.043673 1.66 0.0307410.12 0.040770 0.64 0.060016 1.16 0.043075 1.68 0.0303450.14 0.044809 0.66 0.059413 1.18 0.042485 1.70 0.0299560.16 0.048292 0.68 0.058791 1.20 0.041903 1.72 0.0295730.18 0.051284 0.70 0.058153 1.22 0.041329 1.74 0.0291970.20 0.053840 0.72 0.057503 1.24 0.040763 1.76 0.0288270.22 0.056010 0.74 0.056842 1.26 0.040206 1.78 0.0284630.24 0.057838 0.76 0.056173 1.28 0.039657 1.80 0.0281050.26 0.059363 0.78 0.055498 1.30 0.039117 1.82 0.0277530.28 0.060619 0.80 0.054819 1.32 0.038584 1.84 0.0274070.30 0.061636 0.82 0.054138 1.34 0.038060 1.86 0.0270670.32 0.062441 0.84 0.053455 1.36 0.037545 1.88 0.0267320.34 0.063058 0.86 0.052772 1.38 0.037037 1.90 0.0264030.36 0.063509 0.88 0.052091 1.40 0.036538 1.92 0.0260790.38 0.063813 0.90 0.051412 1.42 0.036046 1.94 0.0257610.40 0.063985 0.92 0.050736 1.44 0.035563 1.96 0.0254480.42 0.064042 0.94 0.050063 1.46 0.035087 1.98 0.0251400.44 0.063997 0.96 0.049396 1.48 0.034619 2.00 0.0248370.46 0.063861 0.98 0.048734 1.50 0.034159 0.48 0.063646 1.00 0.048077 1.52 0.033706 0.50 0.063361 1.02 0.047427 1.54 0.033261
Plot of fit with ρ0 = −1×10−19 C/fm5 to data (plot of the data is not required, just the fit): )( 2qnF
ctor
1.00
Fit to Neutron Form Fa
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0.00 0.50 1.50 2.00
q^2 (GeV/c)^2
F(q^
2)
Fit Data
Corresponding plot of neutron radial charge distribution for ρ0 = −1×10−19 C/fm5 )(2 rr ρ
r^2*rho (C/fm)
-1.5E-21
-1.0E-21
-5.0E-22
0.0E+00
5.0E-22
1.0E-21
1.5E-21
2.0E-21
2.5E-21
3.0E-21
0 2 4 6 8 1
r (fm)0
r^2*rho (C/fm)
Appendix A: This “Macho” integration process (not required for credit):
[ ])1(12)3(4
0sin)/exp()12(
4)( 2
1022
102 BaBq
edrqrararrq
eqnF −
=
∞
−−
=
−−
∫ hhh
πρπρ
where
∫∞
−≡
0sin)/exp()( drqrarrkB k
h
From the Wikipedia integral table under “Exponential Functions”
http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions,
we have the following indefinite integrals (coefficients changed to avoid confusion) and identifying η → −λ:
( ) ( )xxxxxxdxxx µµµλµλλµµµη
µηηµη cossin)exp(cossin)exp()sin()exp( 2222 +
+−−
=−+
=∫
( ) ( )xxxxxxdxxx µµµλµλλµµµη
µηηµη sincos)exp(sincos)exp()cos()exp( 2222 −
+−−
=++
=∫
We start with B(3) from the first term in the form factor. Integrating by parts, and identifying η → −λ:
∫∫∫∞
∞∞∞
−==−=0
000
3 )sin()exp()3( vduuvudvdxxxxB µλ , where u dxxxdvx )sin()exp(,3 µλ−==
( )xxxdxxxdvv µµµλµλλµλ cossin)exp()sin()exp( 22 +
+−−
=−== ∫∫ , and: 23xdu =
( ) [ ])2()2(3cossin)exp()sin()exp()3( 220
223
0
3 CBxxxxdxxxxB µλµλ
µµµλµλλµλ +
++
+
+−
−=−=∞∞
∫
The boundary ( ∞
0uv ) term vanishes, and . Integrating by parts once more: ∫
∞
−≡0
cos)exp()( xdxxxkC k µλ
( ) [ ])1()1(2cossin)exp()sin()exp()2( 220
222
0
2 CBxxxxdxxxxB µλµλ
µµµλµλλµλ +
++
+
+−
−=−=∞∞
∫
( ) [ ])1()1(2cossin)exp()cos()exp()2( 220
222
0
2 BCxxxxdxxxxC µλµλ
µµµλµλλµλ −
++
+
+−
−=−=∞∞
∫
The boundary ( ∞
0uv ) term in each B(2) and C(2) also vanishes and :
( )( )[ ])1(2)1(6)3( 22
222CBB λµµλ
µλ+−
+=
Which happens to involve the second term in the form factor:
( ) [ ])0()0(1cossin)exp()sin()exp()1( 220
220
CBxxxxdxxxxB µλµλ
µµµλµλλµλ +
++
+
+−
−=−=∞∞
∫
( ) [ ])0()0(1sincos)exp()cos()exp()1( 220
220
BCxxxxdxxxxC µλµλ
µµµλµλλµλ −
++
−
+−
−=−=∞∞
∫
Here again the boundary term vanishes and
( ) 22220
220
)0exp(0cossin)exp()sin()exp()0(µλ
µµλ
µµµµλµλλµλ
+=
++=
+
+−−
=−=∞∞
∫ xxxdxxxB
( ) 22220
220
)0exp(0sincos)exp()cos()exp()0(µλ
λµλ
λµµµλµλλµλ
+=
++=
−
+−
−=−=∞∞
∫ xxxdxxxC
And thus we have:
[ ]( )
( )[ ]22222222222222 )/()/1(
/221)0()0(1)1(h
h
qaaqCBB
+=
+=
+
+++
=++
=µλ
λµµλ
µµµλ
µλµλ
µλµλ
[ ]( ) [ ]222
22
222
22
22222222 )/()/1()/()/1(1)0()0(1)1(
h
h
qaqaBCC
+
−=
+
−=
+
−++
=−+
=µλ
µλµλ
µµµλ
λλµλ
µλµλ
Substituting back into the expression for B(3) we have:
( )( )[ ]
( )( )
( )( )
( )
+
−+
+
⋅−
+=+−
+= 222
22
222
22222
22222
226)1(2)1(6)3(µλ
µλλµµλ
λµµλµλ
λµµλµλ
CBB
( )( )422
22
24)3(µλ
µλλµ
+
−=B
Evaluating the integral in and identifying the substitutions λ → 1/a, µ→ : )( 2qnF h/q
( )( ) ( ) ( )
( )( )
−
+
−
+=
+−
+
−=− 2222
22
222222
2422
222 12421224)1(12)3(
λµλ
µλ
µλ
λµ
µλ
λµ
µλ
µλλµ aBaB
( )( ) ( )
( ) ( ) ( )
+
−−−−
+=
+
+−−
+= 2222
4224224
2222222
222222
222
22424µλλ
µµλλµλλ
µλ
λµ
µλλ
µλµλλ
µλ
λµ
( ) ( )( )
( )[ ]
[ ]422
223
4222
223
2222
422
222 )/()/1()/1()/()/1(3)/(24324324
h
hh
qaaqaq
+
+−=
+
+−=
+
−−
+=
µλλ
µλλµ
µλλ
µµλ
µλ
λµ
23∞
[ ][ ]42
422
)/(1)/(3)/(24
0sin)/exp()12(
h
hh
h qaqaqaadrqrararr
+
+−=
−−⇒ ∫
Appendix B: the A=179 case…also acceptable: : tabulated and plotted )( 2qF
q^2 (GeV/c)^2 F(q^2)
q^2 (GeV/c)^2 F(q^2)
q^2 (GeV/c)^2 F(q^2)
q^2 (GeV/c)^2 F(q^2)
0.1 -1.26E-03 0.33 -2.42E-05 0.56 -5.09E-06 0.79 -1.63E-06 0.11 -2.63E-03 0.34 8.82E-06 0.57 -2.49E-06 0.8 -1.59E-06 0.12 -2.82E-03 0.35 3.38E-05 0.58 -1.92E-07 0.81 -1.49E-06 0.13 -2.29E-03 0.36 5.04E-05 0.59 1.73E-06 0.82 -1.35E-06 0.14 -1.46E-03 0.37 5.93E-05 0.6 3.25E-06 0.83 -1.17E-06 0.15 -6.29E-04 0.38 6.13E-05 0.61 4.34E-06 0.84 -9.73E-07 0.16 4.62E-05 0.39 5.81E-05 0.62 5.04E-06 0.85 -7.66E-07 0.17 4.95E-04 0.4 5.09E-05 0.63 5.37E-06 0.86 -5.60E-07 0.18 7.19E-04 0.41 4.13E-05 0.64 5.38E-06 0.87 -3.61E-07 0.19 7.60E-04 0.42 3.05E-05 0.65 5.13E-06 0.88 -1.75E-07
0.2 6.73E-04 0.43 1.97E-05 0.66 4.67E-06 0.89 -7.77E-09 0.21 5.13E-04 0.44 9.51E-06 0.67 4.06E-06 0.9 1.38E-07 0.22 3.27E-04 0.45 6.47E-07 0.68 3.35E-06 0.91 2.59E-07 0.23 1.48E-04 0.46 -6.57E-06 0.69 2.60E-06 0.92 3.55E-07 0.24 -1.20E-06 0.47 -1.20E-05 0.7 1.85E-06 0.93 4.27E-07 0.25 -1.11E-04 0.48 -1.56E-05 0.71 1.13E-06 0.94 4.75E-07 0.26 -1.79E-04 0.49 -1.76E-05 0.72 4.61E-07 0.95 5.00E-07 0.27 -2.10E-04 0.5 -1.80E-05 0.73 -1.24E-07 0.96 5.06E-07 0.28 -2.10E-04 0.51 -1.73E-05 0.74 -6.16E-07 0.97 4.94E-07 0.29 -1.87E-04 0.52 -1.57E-05 0.75 -1.01E-06 0.98 4.68E-07
0.3 -1.51E-04 0.53 -1.34E-05 0.76 -1.30E-06 0.99 4.30E-07 0.31 -1.08E-04 0.54 -1.07E-05 0.77 -1.50E-06 0.32 -6.39E-05 0.55 -7.88E-06 0.78 -1.61E-06
F(q^2) for A=179
-3.50E-03
-3.00E-03
-2.50E-03
-2.00E-03
-1.50E-03
-1.00E-03
-5.00E-04
0.00E+00
5.00E-04
1.00E-03
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
F(q^2)
) with the vertical scale zoomed in for A=179 (not required). ( 2qF
F(q^2) for A=179
-1.00E-04
-8.00E-05
-6.00E-05
-4.00E-05
-2.00E-05
0.00E+00
2.00E-05
4.00E-05
6.00E-05
8.00E-05
1.00E-04
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
F(q^2)
The differential cross section for electron scattering on gold (A=179) vs. scat. angle in degrees.
ds/dW (fm^2) for A=179
1.00E-19
1.00E-16
1.00E-13
1.00E-10
1.00E-07
1.00E-04
0.00 30.00 60.00 90.00 120.00 150.00 180.00
ds/dW (fm^2)