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

For an elastic head-on (collinear) collision we first recall the standard result from conservation of linear momentum and kinetic energy. If a projectile of mass $$m_1$$ and initial speed $$v_1$$ strikes a stationary target of mass $$m_2$$, its speed just after collision is

$$v_1'=\frac{m_1-m_2}{m_1+m_2}\,v_1.$$

Its initial kinetic energy is $$E_i=\tfrac12\,m_1v_1^{\,2}$$, while the kinetic energy after collision is

$$E_f=\tfrac12\,m_1v_1'^{\,2} =\tfrac12\,m_1\!\left(\frac{m_1-m_2}{m_1+m_2}\right)^{\!2}v_1^{\,2} =E_i\!\left(\frac{m_1-m_2}{m_1+m_2}\right)^{\!2}.$$

The fractional loss of energy is therefore

$$P=\frac{E_i-E_f}{E_i} =1-\left(\frac{m_1-m_2}{m_1+m_2}\right)^{\!2}.$$

We now evaluate this expression for the two given targets, taking the neutron mass as $$m$$.

Neutron-deuterium collision: here $$m_1=m$$ and $$m_2=2m$$. Hence

$$\frac{m_1-m_2}{m_1+m_2}=\frac{m-2m}{m+2m} =\frac{-m}{3m} =-\frac13,$$

so

$$P_d=1-\left(-\frac13\right)^{\!2} =1-\frac19 =\frac89 =0.89\;(\text{approximately}).$$

Neutron-carbon collision: here $$m_2=12m$$, therefore

$$\frac{m_1-m_2}{m_1+m_2}=\frac{m-12m}{m+12m} =\frac{-11m}{13m} =-\frac{11}{13},$$

giving

$$P_c=1-\left(-\frac{11}{13}\right)^{\!2} =1-\frac{121}{169} =\frac{48}{169} \approx0.28.$$

Thus the fractional energy losses are $$P_d \approx 0.89$$ and $$P_c \approx 0.28$$.

Hence, the correct answer is Option B.