A particle of mass 200 MeV/c2 collides with a hydrogen atom at rest… | JEE Main 2020 (5 Sep, Shift 1)

Q53JEE Main 2020 · 5 SEP · SHIFT 1Difficult

Transitions, Photon Emission & Excitation

A particle of mass 200 MeV/c2 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

$\frac{N}{4}$ . The value of N is:

(Given the mass of the hydrogen atom to be 1 GeV/c2) ______.

Numerical answer type. Work it out, then open the answer.

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Answer: 51

Momentum conservation: $mu = Mv \Rightarrow v = \frac{m}{M}u = \frac{u}{5}$ (since $m = 0.2$ GeV $/c^2$ , $M = 1$ GeV $/c^2$ ).
KE of recoiling atom $= \frac{1}{2}Mv^2 = \frac{m}{M}K = \frac{K}{5}$ , where $K$ is the initial KE.
Energy used for excitation: $K - \frac{K}{5} = \frac{4K}{5} = 10.2\text{ eV}$ ( $1 \to 2$ in H).
$K = 12.75\text{ eV} = \frac{51}{4}$ eV.
So $N = 51$ .

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