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ID: past-exam-of-the-sub-atomic-physics-course-of-the-university-of-oxford/2016/3/e/solution

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Past exam of the sub-atomic physics course of the University of Oxford / 2016 / 3 / e / Solution by

Codex  0 2026-10-25

The supplied average retained-energy fraction in one elastic collision is

f = \frac{11.1 8 ^{2} + 0.9 4 ^{2}}{( 11.18 + 0.94 ) ^{2}} = 0.8569149.

The mean-energy neutron moderation count follows by iterating the stated mean-energy model for neutron moderation, giving

E_{n} = f^{n} E_{0}

. Equivalently, take logarithms to solve the geometric decrease:

n = \frac{l n ( E _{n} / E _{0} )}{l n f} = \frac{l n [ 0.025/ ( 2 \times 1 0 ^{6} )]}{l n ( 0.8569149 )} = 117.847.

Hence the estimated integer count is

n = 118 collisions .

After

117

collisions the predicted mean energy is about

0.0285

eV; after

118

it is about

0.0244

eV, bracketing the target. This is a mean-energy estimate using the given average retention factor. Individual scattering angles and energy losses fluctuate, so a specified single neutron is not guaranteed to reach the target on exactly its

118

th collision. An estimate based on the average logarithmic energy decrement is a different statistical convention.

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