Physics 3401, Homework 6
Due Monday, November 2
Rohlf Chapter 6, #11,13,14,15,17,18,24
Example problems: 16,19,20,21,22

Physical constants
Boltzmann's constant  k= 1.38066E-23 J/K
k= 0.00008617 eV/K
Atomic mass unit   u= 1.66E-27 kg
umev= 931.5 MeV/c^2
Wien constant   2.90E-03
Planck's constant  h= 4.14E-15 eV s
h= 6.63E-34 J s
h/ 2 pi hbar 1.05E-34 J s
h/ 2 pi hbev 6.59E-16 eV s
hc hc 1.24E+03 eV nm
Speed of light  c= 3.00E+08 m/s
Electron charge  e= 1.60E-19 C
electron volt  eV= 1.60E-19 joule
electron mass  mel 9.11E-31 kg
mel 5.11E-01 MeV
proton mass  mp 1.67E-27 kg
mpev 9.38E+02 MeV/c^2
neutron mass  mn 1.67E-27 kg
mnev 9.40E+02 MeV/c^2

Stefan's constant=  sigma= 5.67E-08 watt/m^2K^4
Nuclear distance unit  fermi 1.00E-15 m (a femtometer)
cross section unit  barn= 1.00E-28 m^2
barn= 100 fm^2
common combo for scattering  ke^2 1.44 MeV fm

#11 Compare alpha scattering to Davisson-Germer
Compare the alpha wavelength to nuclear size and separation between nuclei.
Compare electron wavelength in Davisson-Germer to atom size and separation between atoms.

The alpha wavelength is comparable to nuclear size, but the separation of atoms is four orders of
magnitude larger, so it can be considered to interact with one nucleus at a time.
In Davisson-Germer, the wavelength of the electron was on the order of the spacing
between atomic layers, so it interacted with multiple atoms.

13.Beam of protons with pc=200 MeV scattering off of aluminum nuclei.
a. Cross-section for scattering angles greater than 10 degrees.
The electron kinetic energy is needed. For pc=200 MeV
KE=sqrt((200 MeV)^2+938.27^2) - 938.27 =    21.07904644 MeV
For aluminum with Z=13, the cross section is given by
sigma= pi b^2 =.5* pi(Zke^2/2 KE)^2(1 + cos theta)/(1 - cos theta)
= .5*pi*(13*1.44 MeV fm)/(21 MeV))^2*(1 + cos 10)/(1-cos 10) =
= 163.0757602 fm^2 or 1.630757602  barns

b. For foil 1E-6 m thick, what fraction scattered at greater than 10 degrees.
The fraction scattered is
Rs/Ri = sigma* Avogadro's number*pathlength*density/(atomic mass*1E-3 kg) p176 Rohlf
= (1.63E-30*6E23*1E-6*2.7E3)/(27*1E-3) =    0.00000978

#14. What thickness of air would have the same cross-section as a 0.2E-6 meter thick gold foil?
The scattering rate is proportional to the pathlength times density divided by atomic mass number.
The density of air is about 1.2 kg/m^3 while that of gold is 19000 kg/m^3.
The average mass number for air is about 29, whereas that for gold is 197

The pathlength of air to be equivalent to the foil is then
Air length equivalent = (2E-7 m)(19E3 kg/m^3)(29)/(1.2 kg/m^3)(197) =     0.000466159
So about .47 mm of air is equivalent in alpha scattering to the gold foil,
so you would have to perform the experiment in vacuum.

#15. Impact parameter and closest approach for a 10 MeV alpha scattering at 90 degrees.
a. The impact parameter for an angle theta is
b=(kq1q2/mv^2)*(1 + cos theta)/(1 - cos theta) but for 90 degrees
b= (2)(47)(1.44 MeV fermi/20 MeV =   6.768 fermi

b. The closest approach is given by
rm = Zke^2/KE + .5*sqrt((2Zke^2/KE)^2 + 4b^2) =
=47*1.44/10 + sqrt((2*47*1.44/10)^2 +4b^2) =    16.33939739 fermi

an alternate approach to the closest approach is given by Fowles and Cassady, Analytical Mechanics

rm= b cos(theta/2)/(1 - sin(theta/2)) =   16.33939739 fermi

#17. Calculate alpha incident rate from the scattering between 60 and 90 degrees.
Alpha particle energy is given as 5 MeV and the scattering rate as one per minute.
The scattering is off silver, which has atomic mass A=47
The cross section is given by
sigma = pi b2^2 - pi b1^2= pi(kq1q2/2 Ke)^2(T1 - T2)
where T2= (1+cos 90)/(1-cos 90)=1
T1=(1+cos60)/(1-cos 60) = 3
sigma = pi*(2*47*1.44 MeV fm/(2*5 MeV))^2*(3-1)
sigma = Rs/Ri= 1151.225921 fm^2 11.51225921 barns
After calculating the cross section, the incident flux can be calculated.
Ri = Rs A(10^-3 kg)/(sigma *L* Avogadros's number*density)
Ri =(1/60)*107.9*1E-3/(11.51E-28*1E-6*6.02E23*1.05E4) =    247.1775825 1/s

#18. Cross section for 5 MeV alpha to be scattered from platinum between 5 and 10 degrees.
The common term for the impact parameters is
Zke^2/KE = 78*1.44MeV fm/5 MeV =    22.464 fm
the angular term (1+cos theta)/(1-cos theta) =   130.6460956 for 10 degrees
524.5824763 for 5 degrees
The net cross section is then given by
sigma = pi*(22.46)^2(524.6-130.6) =   624525.4544 fm^2 = 6245.254544

#24. Scattered 10 GeV electrons from protons at 30 degrees. Find energy of scattered electrons.
This is like Compton scattering, so that the scattered energy is given by
KE2 =KE1 m0c^2/(m0c^2 + KE1(1-cos theta))

KE2 = 10 GeV*0.94 GeV/(0.94 GeV + 10 GeV*(1 - cos 30)) =    4.129296308 GeV

Example problems:
#16. Kinetic energy of alpha particle if closest approach is 10 fm when scattered at 90 degrees.
For this angle the impact parameter is  just b = Zke^2/KE
The quadratic equation for the closest approach is
r^2 -2*b*r -b^2, so this gives b =

#19. Find cross section and angle if fraction is 1E-3

#20. For thin sheet of iron, find largest value of KE for which Rutherford is followed.

#21. Kinetic energy for departure from Rutherford on copper target

#22. For 6 MeV alphas, for what Z do you begin to see departure from Rutherford?