If a 2.0-MeV neutron released in a fission reaction loses half its kinetic energy in each moderator collision,how many collisions are needed to reduce its kinetic energy to 1/25 eV?
A)6
B)18
C)26
D)30
E)4

Question

Quizplus · Accepted Answer

To reduce the kinetic energy of the neutron from 2.0 MeV to 1/25 eV, we need to divide the initial energy by (2.0 MeV) / (1/25 eV) = 80,000. If the neutron loses half its kinetic energy in each collision, then the number of collisions needed to reduce its energy by a factor of 80,000 is log2(80,000) = 16.6, or approximately 17. However, each collision reduces the kinetic energy of the neutron by a factor of two, so we actually need twice as many collisions, or 34 collisions, to reduce the neutron's energy by a factor of 80,000.

Therefore, the correct answer is closest to C (26).

If a 20-MeV Neutron Released in a Fission Reaction Loses Half Its