PART 1/2

A neutron in a reactor makes an elastic headon collision with the nucleus of an atom initially at rest.

Assume: The mass of the atomic nucleus is

about 14.1 the mass of the neutron.

What fraction of the neutron's kinetic energy is transferred to the atomic nucleus?

PART 2/2

If the initial kinetic energy of the neutron is

6.98 * 10-13 J, find its final kinetic energy.

Answer in units of J.

0.247

5.25*10⁻¹³ J

Explanation:

Part 1/2

Elastic collision means both momentum and energy are conserved.

Momentum before = momentum after

m v = m v₁ + 14.1m v₂

v = v₁ + 14.1 v₂

Energy before = energy after

½ m v² = ½ m v₁² + ½ (14.1m) v₂²

v² = v₁² + 14.1 v₂²

We want to find the fraction of the neutron's kinetic energy is transferred to the atomic nucleus.

KE/KE = (½ (14.1m) v₂²) / (½ m v²)

KE/KE = 14.1 v₂² / v²

KE/KE = 14.1 (v₂ / v) ²

We need to find the ratio v₂ / v. Solve for v₁ in the momentum equation and substitute into the energy equation.

v₁ = v - 14.1 v₂

v² = (v - 14.1 v₂) ² + 14.1 v₂²

v² = v² - 28.2 v v₂ + 198.81 v₂² + 14.1 v₂²

0 = - 28.2 v v₂ + 212.91 v₂²

0 = - 28.2 v + 212.91 v₂

28.2 v = 212.91 v₂

v₂ / v = 28.2 / 212.91

v₂ / v = 0.132

Therefore, the fraction of the kinetic energy transferred is:

KE/KE = 14.1 (0.132) ²

KE/KE = 0.247

Part 2/2

If a fraction of 0.247 of the initial kinetic energy is transferred to the atomic nucleus, the remaining 0.753 fraction must be in the neutron.

Therefore, the final kinetic energy is:

KE = 0.753 (6.98*10⁻¹³ J)

KE = 5.25*10⁻¹³ J