Q 11-05-201JEE MainJEE Main 2018 (8 Apr)Medium
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:
Answer: (B) $0.89,\ 0.28$
For a head-on elastic collision of mass $m$ with mass $M$ at rest, $v' = \dfrac{m - M}{m + M}v$, so the fractional loss of kinetic energy is
$$P = 1 - \left(\frac{m - M}{m + M}\right)^2$$
- Deuterium ($M = 2m$): $P_d = 1 - \left(\dfrac13\right)^2 = \dfrac89 \approx 0.89$
- Carbon ($M = 12m$): $P_c = 1 - \left(\dfrac{11}{13}\right)^2 = \dfrac{48}{169} \approx 0.28$
Solution by Sreeraj P, M.Sc Physics