Moderators. Canadian nuclear reactors use heavy water moderators in which elasti

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 1/6560 of its original value?

Explanation / Answer

First, conserve momentum:
m*U + 0 = m*V + 2m*v
where V, v are the post-collision velocity of neutron, deuteron
m cancels, so

U = V + 2v For an elastic, head-on collision, we know (from CoE) that
the relative velocity of approach = relative velocity of separation, or
U = v - V
v = U + V plug that into momentum equation

U = V + 2(U + V) = 2U + 3V
U = -3V
V = -U / 3
Speed isn't negative, so the speed ratio is 1/3 = 0.33

B) If the speed is 1/3, the KE is 1/9

C)

(1/3) = 1/6560
3 = 6560
nln3 = ln6560  
n = ln6560 / ln3 = 8 ............answer

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