A neutron in a nuclear reactor makes an elastic, head-on collision with the nucleus of a carbon atom initially at rest.

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.)

 

(b) The initial kinetic energy of the neutron is 1.90  10^-13 J. Find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision.

neutron	J
carbon nucleus	J

Answer 1

Let

m (unknown) be the mass of the neutron,

M = 12m be the mass of the carbon nucleus,

v₀ (unknown) be the initial velocity of the neutron,

v₁ (unknown) be the final velocity of the neutron,

w₀ = 0 be the initial velocity of the carbon nucleus,

w₁ (unknown) be the final velocity of the carbon numcleus.

In the collision momentum is conserved:

mv₀ + Mw₀ = mv₁ + Mw₁ (1)

Since it is an elastic collision, kinetic energy is also conserved:

mv₀²/2 + Mw₀²/2 = mv₁²/2 + Mw₁²/2 (2)

After using M = 12m and w₀ = 0, (1) can be simplified into

v₀ = v₁ + 12w₁ (3)

After using M = 12m and w₀ = 0, (2) can be simplified into

v₀² = v₁² + 12w₁² (4)

From (3)

v₁ = v₀ - 12w₁ (5)

Substitute (5) into (4)

v₀² = (v₀ - 12w₁)² + 12w₁²

This can be simplified into

156w₁² - 24v₀w₁ = 0

This equation has two solutions: w₁ = 0 and w₁ = 24v₀/156 = 2v₀/13

The solution w₁ = 0 corresponds to the neutron just passing through the carbon nucleus,

which we reject and consider only the solution w₁ = 2v₀/13.

We are to calculate the ratio between the final kinetic energy of the carbon nucleus

and the initial kinetic energy of the neutron.

That is, the ratio

(Mw₁²/2) / (mv₀²/2) = (M/m)(w₁/v₀)² = 12×(2/13)² = 0.284

(b)

The final kinetic energy of the carbon nucleus is 1.90×10^-13×0.284 = 0.54×10^-13 J

The final kinetic energy of the neutron is the remainder

1.90×10^-13 - 0.54×10^-13 = 1.36×10^-13 J