Question 2.15: What is the minimum number of elastic scattering collisions ......
What is the minimum number of elastic scattering collisions required to slow a neutron down from 1.0 MeV to 1.0 eV in
a. deuterium,
b. carbon-12,
c. iron-56, and
d. uranium-238?
The minimum number of collision will result if the neutron looses the maximum amount of energy with each collision; That is From Eq. (2.46) E =aE^′. Thus we have 1.0e V= \alpha ^{N}10^{6}e V where N is the minimum number of collisions. Solving we have N=-\ln(10^{6})/\ln(\alpha)=-13.8/\ln(\alpha)\;. Thus with \alpha=(A-1)^{2}\,/(A+1)^{2}
a. deuterium, \alpha = 0.111 ln(\alpha) = – 2.20 N = 6.28 → 7
b. carbon-12, \alpha = 0.716 ln(\alpha) = -0.334 N = 41.3 → 42
c. iron-56, and \alpha = 0.931 ln(\alpha) = -0.0714 N = 193.2 → 194
d. uranium-238? \alpha = 0.983 ln(\alpha) = -0.0168 N = 821.1 → 822