Physics 5110 HW #04 Solutions 1. Ajui/5110/sol/hw04sol.pdf · The two following plots are NOT...

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)




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


222

0

2 CBxxxxdxxxxB µλµλ


++

+

+−

−=−=∞∞

∫

( ) [ ])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+−

+=

http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions

Which happens to involve the second term in the form factor:


220

CBxxxxdxxxxB µλµλ


++

+

+−

−=−=∞∞

∫

( ) [ ])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)

Physics 5110 HW #04 Solutions 1. Ajui/5110/sol/hw04sol.pdf · The two following plots are NOT...

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Transcript of Physics 5110 HW #04 Solutions 1. Ajui/5110/sol/hw04sol.pdf · The two following plots are NOT...