1. Formula for 1D elastic head-on collision

For a particle of mass $m$ and initial speed $u$ colliding elastically and collinearly with a stationary target of mass $M$, the final speed of the incident particle is

$v = \frac{m-M}{m+M}u$

So its final kinetic energy is

$K_f = \frac{1}{2}m v^2 = \frac{1}{2}m\left(\frac{m-M}{m+M}\right)^2 u^2$

If initial kinetic energy is

$K_i = \frac{1}{2}mu^2,$

then the fraction of energy retained is

$\frac{K_f}{K_i} = \left(\frac{m-M}{m+M}\right)^2$

Hence, the fractional loss of energy is

$p = 1 - \left(\frac{m-M}{m+M}\right)^2$

2. Collision of neutron with deuterium

For neutron,

$m = 1$

For deuterium nucleus,

$M = 2$

Thus,

$p_d = 1 - \left(\frac{1-2}{1+2}\right)^2$

$= 1 - \left(\frac{-1}{3}\right)^2$

$= 1 - \frac{1}{9} = \frac{8}{9} \approx 0.89$

3. Collision of neutron with carbon nucleus

For carbon nucleus,

$M = 12$

Thus,

$p_c = 1 - \left(\frac{1-12}{1+12}\right)^2$

$= 1 - \left(\frac{-11}{13}\right)^2$

$= 1 - \frac{121}{169}$

$= \frac{48}{169} \approx 0.284 \approx 0.28$

4. Match with options

So,

$p_d \approx 0.89, \qquad p_c \approx 0.28$

This matches Option B.