It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is $${P_d};$$ while for its similar collision with carbon nucleus at rest, fractional loss of energy is $${P_c}.$$ The values of $${P_d}$$ and $${P_c}$$ are respectively:

Solution :
For collision of neutron with deuterium:

Applying conservation of momentum :
$$\eqalign{ & mv + 0 = m{v_1} + 2m{v_2}\,......\left( {\text{i}} \right) \cr & {v_2} - {v_1} = v\,......\left( {{\text{ii}}} \right) \cr} $$
$$\because $$ Collision is elastic, $$e = 1$$
From eqn (i) and eqn (ii) $${v_1} = - \frac{v}{3}$$
$${P_d} = \frac{{\frac{1}{2}m{v^2} - \frac{1}{2}mv_1^2}}{{\frac{1}{2}m{v^2}}} = \frac{8}{9} = 0.89$$
Now, For collision of neutron with carbon nucleus

Applying Conservation of momentum
$$\eqalign{ & mv + 0 = m{v_1} + 12m{v_2}\,......\left( {{\text{iii}}} \right) \cr & v = {v_2} - {v_1}\,......\left( {{\text{iv}}} \right) \cr} $$
From eqn (iii) and eqn (iv)
$$\eqalign{ & {v_1} = - \frac{{11}}{{13}}v \cr & {P_c} = \frac{{\frac{1}{2}m{v^2} - \frac{1}{2}m{{\left( {\frac{{11}}{{13}}v} \right)}^2}}}{{\frac{1}{2}m{v^2}}} = \frac{{48}}{{169}} \approx 0.28 \cr} $$