Question

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 14.9 the mass of the neutron. what fraction of the neutron's kinetic energy is transferred to the atomic nucleus?

Answer 1

Answer: The fraction of the neutron's kinetic energy transferred to the atomic nucleus is 1.87.

Explanation:

Use the conservation of momentum in head-on collision.

$mv_{i}=mv_{f}+MV_{f}$    

$v_{i}-v_{f}=\frac{M}{m}V_{f}$          

Put

$\frac{M}{m}=14.9$

$v_{i}-v_{f}=14.9V_{f}$                               ....... (1)                                                                                  

Here, m, M are the masses of the neutron and atom, $v_{i}$ is the initial velocity of the neutron,$v_{f}$ is the final velocity of the neutron and $V_{f}$ is the final velocity of the atom.

Use the conservation of the kinetic energy.

$\frac{1}{2}mv_{i}^{2}=\frac{1}{2}mv_{f}^{2}+\frac{1}{2}MV_{f}^{2}$

$v_{i}^{2}-v_{f}^{2}=\frac{M}{m}V_{f}^{2}$

Put

$\frac{M}{m}=14.9$.

$v_{i}^{2}-v_{f}^{2}=14.9V_{f}^{2}$                                   ...... (2)

Divide (2) by (1).

$\frac{v_{i}^{2}-v_{f}^{2}}{v_{i}-v_{f}}=\frac{14.9V_{f}^{2}}{14.9V_{f}}$

$\frac{v_{i}^{2}-v_{f}^{2}}{v_{i}-v_{f}}=V_{f}$

$v_{i}+v_{f}=V_{f}$                                                              ...... (3)

Solve equation (1) and (3).

$v_{i}=7.9V_{f}$

Calculate the fraction of the neutron's kinetic energy transferred to the atomic nucleus.

$Fraction\ transferred\ to\ atom =\frac{\frac{1}{2}MV_{f}^{2}}{\frac{1}{2}mv_{i}^{2}}$

Put

$\frac{M}{m}=14.9$

$Fraction\ transferred\ to\ atom=\frac{14.9}{7.95}$

$Fraction\ transferred\ to\ atom=1.87$

Therefore, the fraction of the neutron's kinetic energy transferred to the atomic nucleus is 1.87.

Answer 2

14.9 more nucleuses is the answer i think