Question

A neutron in a nuclear reactor collides head on elastically with the nucleus of a carbon atom initially at rest. The fraction of kinetic energy transferred from the neutron from the carbon atom is
(A). $\dfrac{11}{12}$
(B). $\dfrac{2}{11}$
(C). $\dfrac{48}{121}$
(D). $\dfrac{48}{169}$

Answer 1

Hint: The neutron collides with the carbon atom which is initially at rest. After the collision, the carbon atom also starts moving as the neutron transfers some of its energy to the carbon atom. From the law of conservation of energy, we can calculate the final velocity of a neutron and carbon atom. The fraction of energy transferred is the change in the kinetic energy of the neutron to the initial kinetic energy.
Formulas used:
$v=\left( \dfrac{m-M}{m+M} \right)u$
$E=\dfrac{\Delta E}{{{E}_{i}}}$

Complete step-by-step solution:
The system of a neutron and the carbon atom is an isolated system therefore; the total momentum of the system will be conserved.
Let the mass of the neutron be $m$, then the mass of the carbon atom will be $12m$.
From the law of conservation of momentum,
Initial momentum of the system is equal to the final momentum of the system.
Let the final velocity of the system of neutrons and carbon be $v$. Therefore, from law of conservation of energy, the final velocity of the system will be-
$v=\left( \dfrac{m-M}{m+M} \right)u$ - (1)
Therefore, the kinetic energy of the neutron initially will be
${{K}_{i}}=\dfrac{1}{2}m{{u}^{2}}$ - (2)
The final kinetic energy of the neutron will be
${{K}_{f}}=\dfrac{1}{2}m{{\left( \dfrac{m-M}{m+M} \right)}^{2}}{{u}^{2}}$ - (3)
The fraction of kinetic energy transferred to the carbon atom from the neutron is the ratio of change in energy of the neutron to the initial energy of the neutron atom. Hence,
$E=\dfrac{\Delta E}{{{E}_{i}}}$
Here, $E$ is the energy transferred
$\Delta {{E}_{C}}$ is the change in the energy of carbon atom
$\Delta {{E}_{N}}$ is the change in the energy of neutron atom
In the above substituting given values we get,
$\begin{align}
  & E=\dfrac{\dfrac{1}{2}m{{u}^{2}}-\dfrac{1}{2}m{{\left( \dfrac{m-M}{m+M} \right)}^{2}}{{u}^{2}}}{\dfrac{1}{2}m{{u}^{2}}} \\
 & \Rightarrow E=\dfrac{{{(m+M)}^{2}}-{{(m-M)}^{2}}}{{{(m+M)}^{2}}} \\
 & \therefore E=\dfrac{4mM}{{{(m+M)}^{2}}} \\
\end{align}$
We know, $M=12m$ substituting in the above equation, we get,
$\begin{align}
  & E=\dfrac{4mM}{{{(m+M)}^{2}}} \\
 & \Rightarrow E=\dfrac{4\times m\times 12m}{{{(13m)}^{2}}} \\
 & \therefore E=\dfrac{48}{169} \\
\end{align}$
Therefore, the fraction of kinetic energy transferred from neutron to the carbon atom is $\dfrac{48}{169}$. Hence, the correct option is (D).

Note: The collision between the neutron and carbon atom is elastic. For an isolated system, the mechanical energy is conserved. The mass of an atom is the sum of the number of protons and neutrons in it, therefore, the carbon has six neutrons and six protons.