(b) In the core of a nuclear reactor, a fission neutron moving at a speed of 3.9 × 106 m s-1 collides with a carbon-12 nucleus which is initially at rest. Immediately after the collision, the carbon nucleus has a velocity of 6.0 × 105 m s-1 in the same direction as the initial direction of the neutron.
(i) Show that the neutron rebounds with a speed of 3.3 × 106 m s-1.
mass of a single carbon 12 nucleus = 0.012/6.02 x 1023
m = 2.0 × 10–26 kg 
By conservation of momentum:
final momentum of neutron and nucleus = initial momentum of neutron
(1.67 × 10–21 × v) + (2.0 × 10–26 × 6.0 ×105) = 1.67 × 10–27 × 3.9 × 106
v =[(1.67 × 10–27 × 3.9 × 106) - (2.0 × 10–26 × 6.0 ×105)]/1.67 × 10–21
v = 3.3 × 106 m s–1
(ii) Show that the collision is an elastic collision.
initial Ek of neutron = ½ × 1.67 × 10–27 × (3.9 × 106)2 = 1.27 × 10–14J
final Ek of neutron = ½ × 1.67 × 10–27 × (3.3 × 106)2) = 9.1 × 10–15J
initial Ek of carbon nucleus = 0 J
final Ek of carbon nucleus = ½ × 2.0 × 10–26 × (6.0 × 105)2) = 3.6 × 10–15J
Total Initial Ek = 1.27 × 10–14 J
Total Final Ek = 9.1 × 10–15J + 3.6 × 10–15J = 1.27 × 10–14 J
(iii) Calculate the percentage of the initial kinetic energy of the neutron that is transferred to the carbon nucleus.
% Ek transferred = (energy gained by nucleus/energy possessed by neutron initially) x 100 % % Ek transferred = (3.6 x 10-15 /1.27 x 10-14) x 100 = 28(.3)%
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