Question

A neutron with velocity $v$ strikes a stationary deuterium atom, its kinetic energy changes by a factor of :

• A. $\frac{15}{16}$
• B. $\frac{1}{2}$
• C. $\frac{2}{1}$
• D. None of these

Answer 1

The correct option is D None of these

Let mass of neutron $=m$

then mass of deuterium $=2m$

[$\because$ it has double nuclides thus has neutron].

Let initial velocity of neutron $=v$ and final velocities of neutron and deuterium are $v_1$ and $v_2$ respectively.

Applying conservation of momentum

$mv+2m\left(0\right)=m{v}_1+2m{v}_2$

$\Rightarrow v=v_1+2v_2$ ...........(i)

applying conservation of energy

$\frac{1}{2}m{v}^2=\frac{1}{2}m{v_1}^2+\frac{1}{2}2m{v_2}^2$

$\Rightarrow v^2={v_1}^2+2{v_2}^2$........(ii)

from (i) and (ii), $v^2=(v-2v_2{)}^2+2{v_2}^2$

$\Rightarrow v^2=v^2+4{v_2}^2-4v_{2}v+2{v_2}^2$

$6{v_2}^2-4v_{2}v=0$

$\Rightarrow v_2=\frac{2v}{3}$ & $v_1=-\frac{v}{3}$

now fractional change in kinetic energy

$=\frac{K_i-K_f}{K_i}=\frac{\frac{1}{2}m{v}^2-\frac{1}{2}m{v_1}^2}{\frac{1}{2}m{v}^2}=\frac{v^2-\frac{v^2}{9}}{v^2}=\frac{8}{9}$

Alternatively,

In one dimensional elastic collision between two particles having masses $m_1$ and $m_2$ having initial velocities $u_1$ and $u_2$ respectively,

$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2$

Substituting $m_1=m$, $m_2=2m$, $u_1=v$, $u_2=0$

$v_1=\frac{-1}{3}v$

$\Rightarrow \frac{K_i-K_f}{K_i}=\frac{\frac{1}{2}m{v}^2-\frac{1}{2}m{v_1}^2}{\frac{1}{2}m{v}^2}=\frac{v^2-\frac{v^2}{9}}{v^2}=\frac{8}{9}$

We have used a general formula rather than applying momentum and energy conservation for the specific condition.