Question

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

We know,

Coefficient of restitution is given as

$e=\frac{v_2-v_1}{v-0}$

$ve=v_2-v_1$

as collision is perfectly elastic $e=1$

$v=v_2-v_1$

using linear momentum conservation,

$mv=m{v}_1+Am{v}_2$

$\Rightarrow (A+1)v_1=(1-A)v$

$v_1=\left(\frac{1-A}{1+A}\right)v$

so, kinetic energy retained $={\left(\frac{1-A}{1+A}\right)}^2$