A neutron in a nuclear reactor makes an elastic, head-on collision with the nucl

Question

A neutron in a nuclear reactor makes an elastic, head-on collision with the nucleus of a carbon atom initially at rest.
(a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? (The mass of the carbon nucleus is about 12.0 times the mass of the neutron.)
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(b) The initial kinetic energy of the neutron is 3.10 10-13 J. Find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision.


neutron

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Your response differs from the correct answer by more than 100%. J



carbon nucleus

Explanation / Answer

For two particles (masses m1 and m2, initial velocities u1 and u2), which undergo a perfectly elastic head-on collision, you can find the after- collision velocities from [1]:
v1 = (u1·(m1 - m2) + 2·m2·u2)/(m1 + m2)
v2 = (u2·(m2 - m1) + 2·m1·u1)/(m1 + m2)

Let subscript 1 denotes the proton and 2 the nucleus. Since the nucleus was initially at rest, i.e. u2=0, the velocity of the nucleus after the collision is:
v2 = 2·m1·u1/(m1 + m2) = 2·u1/(1 + (m2/m1))

So the kinetic energy of the nucleus after the collision is:
E2' = (1/2)·m2·v2² = 2·m2·u1²/(1 + (m2/m1))²

The fraction of the neutrons kinetic energy transferred to the nucleus is:
f = E2'/E1
= [ 2·m2·u1²/(1 + (m2/m1))² ] / [ (1/2)·m1·u1²) ]
= 4·(m2/m1) / (1 + (m2/m1))²
= 4· 12 / (1 + 12)²
= 0.28
= 28%

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