#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.

#11 将α粒子散射与达维斯-格默实验相比较。将α粒子的波长与核尺寸及核之间的距离相比较。将达维斯-格默实验中的电子波长与原子尺寸及原子之间的距离相比较。

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.

α射线的波长与核尺寸相当，但原子间的距离大约大四个数量级，因此可以认为它一次只与一个核相互作用。在德拜-germer实验中，电子的波长与原子层之间的间距相当，因此它与多个原子相互作用。

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

13. 一束动量为200 MeV的质子束散射在铝核上。a. 散射角大于10度时的截面。需要电子的动能。对于pc=200 MeV，动能KE=√((200 MeV)^2+938.27^2) - 938.27 = 21.08 MeV。对于铝（Z=13），截面由sigma=πb²给出，其中sigma=π*(13*1.44 MeV fm/(2*21 MeV))^2*(1 + cos 10°)/(1 - cos 10°) = 81.00 fm²或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

b. 对于厚度为1E-6 m的箔，散射角度大于10度的分数是多少？散射分数为Rs/Ri = sigma*阿伏加德罗常数*路径长度*密度/(原子质量*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.

#14. 什么厚度的空气具有与0.2微米厚的金箔相同的截面？散射率与路径长度、密度、Z²/原子质量成正比。空气的密度约为1.2 kg/m³，而金的密度为19000 kg/m³。那么，与箔片相当的空气路径长度为：Air length equivalent = (2E-7 m)(19000 kg/m³)/(1.2 kg/m³)*(14/197)*(79/7)^2 m Air length equivalent = 0.028663041 m = 29 mm 使用14是将空气视为主要由氮组成，你将会散射到一个N原子，即使在N₂中也是如此。因此，大约29毫米的空气在α散射方面等同于金箔，所以你必须在真空中进行实验。

#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

#15. 10 MeV alpha粒子在90度散射时的冲击参数和最近接近距离。a. 对于角度theta，冲击参数b=(kq1q2/mv²)*(1 + cosθ)/(1 - cosθ)，但当角度为90度时，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

b. 最近距离由rm = Zke²/KE + 0.5*sqrt((2Zke²/KE)² + 4b²) = =47*1.44/10 + sqrt((2*47*1.44/10)² +4b²) = 16.34 佛米

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

另一种方法见Fowles和Cassady的《分析力学》

rm= b cos(theta/2)/(1 - sin(theta/2)) = 16.34 佛米

#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

#17. 从60到90度之间的散射计算α粒子的入射率。已知α粒子的能量为5 MeV，散射率为每分钟1次。散射发生在银上，银的原子量A=47。散射截面由sigma = πb2² - πb1² = π(kq1q2/2Ke)²(T1 - T2)给出，其中T2=(1+cos90)/(1-cos90)=1，T1=(1+cos60)/(1-cos60)=3。sigma = π*(2*47*1.44 MeV fm/(2*5 MeV))²*(3-1)。sigma = Rs/Ri= 1151.225921 fm² 11.51 barns。在计算出散射截面后，可以计算入射通量。Ri = Rs A(10^-3 kg)/(sigma *L* Avogadro's number*density)。Ri =(1/60)*107.9*1E-3/(11.51E-28*1E-6*6.02E23*1.05E4) = 247.18 1/s #18. 5 MeV的α粒子散射从铂的5到10度之间的散射截面。冲击参数的公共项是Zke²/KE = 78*1.44MeV fm/5 MeV = 22.464 fm。角度项(1+cosθ)/(1-cosθ) = 130.65对于10度，524.58对于5度。因此，净散射截面由sigma = π*(22.46)²(524.6-130.6) = 624525.45 fm² = 6245.25 barns给出。

#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))

24. 从质子中散射出的10 GeV电子，在30度角处散射。求散射电子的能量。这类似于康普顿散射，因此散射后的能量由公式 KE2 = KE1 m0c^2/(m0c^2 + KE1(1 - cos theta)) 给出。