Question

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

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

Answer 1

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

Let mass of the neutron be $m$.

Velocity of neutron before collision $u_1=u$

So, its initial kinetic energy $E=\frac{1}{2}m{u}^2$

Mass of the nucleus $M=mA$

Its velocity before collision $u_2=0$

Velocity of neutron just after collision $v_1=\frac{(m_1-e{m}_2)u_1+(1+e)m_2{u}_2}{m_1+m_2}$

where $m_1=m$, $m_2=mA$, $u_2=0$, $e=1$ and $u_1=u$

$\therefore$ $v_1=\frac{(m-mA)u+(1+e)\left(mA\right)\left(0\right)}{m+mA}=\frac{(1-A)}{(1+A)}u$

Thus kinetic energy of neutron after collision $E^{\prime }=\frac{1}{2}m{v}_1^2=\frac{1}{2}m{u}^2(\frac{1-A}{1+A}{)}^2$

So, fraction of kinetic energy retained by neutron $\frac{E^{\prime }}{E}=(\frac{1-A}{1+A}{)}^2$