Q 12-12-108JEE MainJEE Main 2020 (5 Sep, Shift 1)Medium
A particle of mass $200\ \text{MeV}/c^{2}$ collides with a hydrogen atom at rest. Soon after the collision, the particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eV) is $\dfrac N4$. The value of $N$ is: (Given the mass of the hydrogen atom to be $1\ \text{GeV}/c^{2}$) ______
Numerical value type. Enter your answer.
Answer: 51
Momentum: $mu = MV$ with $\dfrac mM = \dfrac15$, so $V = \dfrac u5$ and the atom's kinetic energy is $K\dfrac mM = \dfrac K5$.
Energy: $K = \dfrac K5 + 10.2$ eV (excitation $n = 1 \to 2$).
$$\frac{4K}{5} = 10.2 \Rightarrow K = 12.75\ \text{eV} = \frac{51}{4}\ \text{eV}$$
So $N = 51$.
Solution by Sreeraj P, M.Sc Physics