Due Monday, November 8

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 the pathlength times density.

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)*
= 0.000225042 m

Using 14 is treating air as mostly nitrogen, and you would scatter
off one N, even in N2 0.225 mm

So about .2 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