Answer:
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
