JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Modern Physics

Momentum is conserved and the particle stops dead, so all its momentum passes to the atom: $M_0{V}_0=mV$ gives the recoil speed $V=\frac{M_0}{m}{V}_0$. Energy conservation then splits the particle's kinetic energy between the atom's recoil energy and the $10.2$ eV needed to raise hydrogen to its first excited state: $\frac{1}{2}{M}_0{V}_0^2=\frac{1}{2}m{V}^2+10.2=\frac{M_0^2{V}_0^2}{2m}+10.2$. Hence $\frac{1}{2}{M}_0{V}_0^2\left(1-\frac{M_0}{m}\right)=10.2$ eV, and with $\frac{M_0}{m}=\frac{200}{1000}=\frac{1}{5}$, $\frac{1}{2}{M}_0{V}_0^2=\frac{10.2}{0.8}=12.75$ eV $=\frac{51}{4}$ eV, so $N=51$.