Question 7.4: In the earlier discussion, we discovered that the average ve......
In the earlier discussion, we discovered that the average velocity of a prompt neutron when it is initially emitted from a uranium nucleus is 9000 times greater than the velocity of the same neutron when it reaches thermal equilibrium. If the most probable energy that a prompt neutron is emitted from a Uranium-235 nucleus is 0.72 MeV, how many times greater is this than the velocity of the neutron when it becomes a thermal neutron?
Since E/E_T = (v/v_T)^2, we know that v/v_T = \sqrt{(E/E_T )}. If the most probable energy of the neutron when it is
emitted is E = 0.72 MeV and the energy of the neutron at thermal equilibrium is 0.025 eV, then E/E_T = 28, 800, 000, and v_{MP} = \sqrt{28,800,000v_T = 5,366v_T}. So the most probable velocity that the neutron is emitted with v_{MP} is about 5366 times the final thermal equilibrium velocity v_T. By any objective measure, this is an enormous difference in the speed.