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

Question

Quizplus · Accepted Answer

We can use the formula for energy loss in a moderator collision:

E_final = E_initial/2^n, where n is the number of collisions.

We want to find n when E_final = (1/25)eV and E_initial = 2.0 MeV = 2.0 x 10^6 eV.

(1/25)eV = (2.0 x 10^6 eV)/2^n

Simplifying, we get:

2^n = (2.0 x 10^6)/(1/25) = 50 x 10^6

Taking the logarithm of both sides:

n log(2) = log(50 x 10^6)

n = log(50 x 10^6)/log(2) ≈ 25.8

Therefore, we need approximately 26 collisions to reduce the energy of the neutron to (1/25)eV. The closest answer choice is C, 26.

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