Question

Moderators. Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons and deuterons of mass 2.0 u.

(a) What is the speed of a neutron, expressed as a fraction of its original speed, after a head-on, elastic collision with a deuteron that is initially at rest?


(b) What is its kinetic energy, expressed as a fraction of its original kinetic energy?


(c) How many such successive collisions will reduce the speed of a neutron to 1/2187 of its original value?

Answer 1

Solving these equations was accomplished before in three dimensions and shown to yield equation (3.36), from which we use only the x-component mi-in md 1.0 u- 2.0 u nx,l 1.0 u 2.0 u ' nr,l (b) The neutron kinetic energy after the collision is nx,l (c) For each collision the speed of the neutron is reduced by a factor of three. So for n collisions we have Lr 2187 This implies that 3° = 2187 OI ln(3" ) = nln(3) In(2187) and ln(21871 In(3) -7 (Answer)