A neutron travelling with a velocity v and K.E. E 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

The correct option is (A-1/A+1)².

To find the fraction of kinetic energy retained by the neutron after a perfectly elastic head-on collision with a nucleus at rest, we can use the principles of conservation of linear momentum and conservation of kinetic energy.

1. Identify the masses and initial velocities:
Let the mass of the neutron be _m1 = m.
Since the mass number of the nucleus is A, its mass is approximately _m2 = Am.
Initial velocity of the neutron, _u1 = v.
Initial velocity of the nucleus, _u2 = 0 (since it is at rest).

2. Formula for final velocity in a one-dimensional elastic collision:
The final velocity of the first body (neutron) after a perfectly elastic collision is given by:
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2$

Substituting the values of masses and initial velocities:
$v_1=\left(\frac{m-Am}{m+Am}\right)v+0$

Simplifying the velocity expression by factoring out m:
$v_1=\left(\frac{1-A}{1+A}\right)v$

3. Calculate kinetic energy retained:
The initial kinetic energy of the neutron (E) is:
$E=\frac{1}{2}m{v}^2$

The final kinetic energy of the neutron (E') is:
$E^{\prime }=\frac{1}{2}m{v}_1^2$

Substitute the value of _v1 into the final kinetic energy equation:
$E^{\prime }=\frac{1}{2}m{\left(\frac{1-A}{1+A}v\right)}^2$

Using the property that (1 - A)² = (A - 1)²:
$E^{\prime }={\left(\frac{A-1}{A+1}\right)}^2\left(\frac{1}{2}m{v}^2\right)$

Therefore, we have:
$E^{\prime }={\left(\frac{A-1}{A+1}\right)}^{2}E$

4. Determine the fraction of total energy retained:
The fraction of the neutron's initial energy that is retained is:
$\frac{E^{\prime }}{E}={\left(\frac{A-1}{A+1}\right)}^2$