Question

A neutron travelling with a velocity $v$ and kinetic energy $E$ has a perfectly elastic head-on collision with a nucleus of an atom of mass number $A$ at rest. The fraction of total energy retained by the neutron is approximately :

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

Answer 1

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

After collision, let the neutron and the nucleus move with speed $v_1$ and $v_2$ respectively.

Mass of the neutron and the nucleus is $m$ and $Am$ respectively where $m\approx m_{proton}\approx m_{neutron}$

As the linear momentum is conserved before and after the collision i.e $P_{initial}=P_{final}$

$mv+0=m{v}_1+\left(Am\right)v_2⟹v=v_1+A{v}_2$ .........(1)

Also total energy is conserved i.e $E_{initial}=E_{Final}$

$\therefore \frac{1}{2}m{v}^2+0=\frac{1}{2}m{v}_1^2+\frac{1}{2}\left(Am\right)v_2^2$

$⟹v^2=v_1^2+A{v}_2^2$ ................(2)

Solving (1) and (2) we get, $v_1=(\frac{A-1}{A+1})v$

Energy of the neutron before collision $E=\frac{1}{2}m{v}^2$

Energy of the neutron after collision $E^{\prime }=\frac{1}{2}m{v}_1^2=\frac{1}{2}m{v}^2(\frac{A-1}{A+1}{)}^2$

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