Question

A neutron travelling with a velocity $v$ and kinetic energy $E$ collides elastically head on with the nucleus of an atom of mass number $A$ at rest. The fraction of total energy retained by the neutron is:

• A. ${\left(\frac{A-1}{A+1}\right)}^2$
• B. ${\left(\frac{A+1}{A-1}\right)}^2$
• C. ${\left(\frac{A-1}{A}\right)}^2$
• D. ${\left(\frac{A+1}{A}\right)}^2$

Answer 1

The correct option is A ${\left(\frac{A-1}{A+1}\right)}^2$

Let, $m_1$ be mass of the neutron and its initial velocity be $u_1$

$m_2$ be mass of the nucleus and its velocity $u_2$ is $0$ since it was at rest initially.

According, to the law of conversation of energy.

$v_1=\frac{(m_1-m_2)u_1+2m_2{u}_2}{m_1+m_2}$

$\Rightarrow v_1=\frac{(m_1-m_2)u_1}{m_1+m_2}(\because u_2=0)$

$\Rightarrow \frac{v_1}{u_1}=\frac{m_1-m_2}{m_1+m_2}$

Taking $m_2$ common and cancelling if, we get

$\Rightarrow \frac{v_1}{u_1}=\frac{1-\frac{m_2}{m_1}}{1+\frac{m_2}{m_1}}$

$\Rightarrow \frac{v_1}{u_1}=\frac{1-A}{1+A}$

$A=\frac{m_2}{m_1}=$ mass no. of nucleus

we know that,

$K.E=\frac{1}{2}m{v}^2$

Fraction of total KE $=\frac{\frac{1}{2}m{v}_1^2}{\frac{1}{2}m{u}_1^2}$

$\Rightarrow \frac{v_1^2}{u_1^2}$

$\therefore$ fraction of total energy retained

$\Rightarrow \frac{v_1^2}{u_1^2}=\frac{(1-A{)}^2}{(1+A{)}^2}$

$={\left(\frac{A-1}{A+1}\right)}^2$