Question

A neutron makes a head on elastic collision with a stationery deuteron. The fractional energy loss of the neutron in the collision is:
(A). $\dfrac{16}{82}$
(B). $\dfrac{8}{9}$
(C). $\dfrac{8}{27}$
(D). $\dfrac{2}{3}$

Answer 1

Hint: The neutron and deuteron undergo elastic collision. Elastic collisions follow the laws of conservation of energy as well momentum. Using the conservation equations, we can substitute the velocities of particles in terms of initial velocity of neutrons. The fractional loss in energy is the loss in kinetic energy divided by initial kinetic energy.

Formula used:
${{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$
$\dfrac{1}{2}{{m}_{1}}u_{1}^{2}+\dfrac{1}{2}{{m}_{2}}u_{2}^{2}=\dfrac{1}{2}{{m}_{1}}v_{1}^{2}+\dfrac{1}{2}{{m}_{2}}v_{2}^{2}$

Complete answer:
A neutron is an atomic particle which does not have a charge on it.
A deuteron has a proton and a neutron bind together. It is also called a deuterium and is the isotope of hydrogen.
The mass of a deuteron is two times the mass of a neutron.
$\begin{align}
  & {{m}_{n}}=m \\
 & {{m}_{d}}=2m \\
\end{align}$
When elastic collision takes place, the momentum as well as energy is conserved. Therefore,
${{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$ - (1)
$\dfrac{1}{2}{{m}_{1}}u_{1}^{2}+\dfrac{1}{2}{{m}_{2}}u_{2}^{2}=\dfrac{1}{2}{{m}_{1}}v_{1}^{2}+\dfrac{1}{2}{{m}_{2}}v_{2}^{2}$ - (2)
Here ${{m}_{1}},\,{{m}_{2}}$ are the masses of neutron and deuteron respectively
${{u}_{1}},\,{{u}_{2}}$are their initial velocities
${{v}_{1}},\,{{v}_{2}}$are their final velocities
Let ${{u}_{1}}=u$, given ${{u}_{2}}=0$
Substituting values in eq (1), we get,
$mu=m{{v}_{1}}+2m{{v}_{2}}$
${{v}_{1}}=\dfrac{mu-2m{{v}_{2}}}{m}$ - (3)
In eq (2), we substitute value of ${{v}_{1}}$to get,
$\begin{align}
  & m{{u}^{2}}=mv_{1}^{2}+2mv_{2}^{2} \\
 & \Rightarrow m{{u}^{2}}=m{{\left( \dfrac{mu-2m{{v}_{2}}}{m} \right)}^{2}}+2mv_{2}^{2} \\
 & {{u}^{2}}={{(u-{{v}_{2}})}^{2}}+2v_{2}^{2} \\
 & 4u=6{{v}_{2}} \\
 & \therefore \dfrac{2}{3}u={{v}_{2}} \\
\end{align}$
$\dfrac{2}{3}u={{v}_{2}}$ - (4)
From eq (3) and eq (4), we get,
$\therefore {{v}_{1}}=-\dfrac{1}{3}u$
The negative sign indicates that the velocity${{v}_{1}}$ is in the direction opposite to$u$.
The fractional loss of energy for the neutron is given as-
$\dfrac{{{E}_{in}}-{{E}_{f}}}{{{E}_{in}}}=\dfrac{\dfrac{1}{2}m{{u}^{2}}-\dfrac{1}{2}mv_{1}^{2}}{\dfrac{1}{2}m{{u}^{2}}}$
$\Rightarrow \dfrac{\Delta E}{E}=\dfrac{{{u}^{2}}-\dfrac{{{u}^{2}}}{9}}{{{u}^{2}}}$
$\therefore \dfrac{\Delta E}{E}=\dfrac{8}{9}$

Therefore, the fractional loss in energy of a neutron after collision is $\dfrac{8}{9}$. So, the correct option is (B).

Note:
Here the neutron changes its direction after collision with the deuteron. The system is isolated as there are no external forces acting on the system. There are no electrostatic forces acting between the neutron and deuteron as there is no charge on the neutron. There is a loss in kinetic energy of the neutron because it transfers some of its energy to the deuteron for it to start moving.