Physics 3401, Homework 6

Rohlf Chapter 6, #11,13,14,15,17,18,24

#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 =    21.08 MeV
For aluminum with Z=13, the cross section is given by
sigma= pi b^2 = pi(Zke^2/2 KE)^2(1 + cos theta)/(1 - cos theta)
= pi*(13*1.44 MeV fm)/(2*21 MeV))^2*(1 + cos 10)/(1-cos 10) =
= 81.00 fm^2 or 0.810 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
= (81E-30*6E23*1E-6*2.7E3)/(27*1E-3) =    4.86E-06

#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 pathlength*density*Z^2/atomic mass
The density of air is about 1.2 kg/m^3 while that of gold is 19000 kg/m^3.
The pathlength of air to be equivalent to the foil is then
Air length equivalent = (2E-7 m)(19E3 kg/m^3)/(1.2 kg/m^3)*(14/197)*(79/7)^2      m
Air length equivalent =  0.028663041 m = 29  mm
Using 14 is treating air as mostly nitrogen, and you would scatter off one N, even in N2      mm
So about 29 mm of air is equivalent in alpha scattering to the gold foil,
so you would definitely 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) =    16.34 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)) =   16.34 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 11.51 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) =    247.18 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) =   130.65 for 10 degrees
524.58 for 5 degrees
The net cross section is then given by
sigma = pi*(22.46)^2(524.6-130.6) =   624525.45 fm^2 = 6245.25 barns

#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)) =    4.13 GeV