1. Given data

• Mass of incident particle:
  $m = 200\ \text{MeV}/c^2$
• Mass of hydrogen atom:
  $M = 1\ \text{GeV}/c^2 = 1000\ \text{MeV}/c^2$
• Hydrogen atom is initially at rest.
• After collision:
  • incident particle comes to rest,
  • hydrogen atom recoils,
  • hydrogen atom goes to first excited state.

For hydrogen, the first excitation energy is: $\Delta E = 10.2\ \text{eV}$

We need the initial kinetic energy of the particle.

2. Apply momentum conservation

Let the initial momentum of the particle be $p$.

Initially:

• particle has momentum $p$
• hydrogen atom has momentum $0$

Finally:

• particle comes to rest
• hydrogen atom recoils with momentum $p$

So by conservation of momentum, recoil momentum of hydrogen atom is also $p$.

3. Apply energy conservation

Initial kinetic energy of particle = final recoil kinetic energy of hydrogen atom + excitation energy.

Thus, $K = \frac{p^2}{2m}$ for the incident particle, and $K_H = \frac{p^2}{2M}$ for the recoiling hydrogen atom.

Hence, $\frac{p^2}{2m} = \frac{p^2}{2M} + \Delta E$

Substitute $\Delta E = 10.2\,\text{eV}$: $\frac{p^2}{2m} - \frac{p^2}{2M} = 10.2$

Factor out $\frac{p^2}{2}$: $\frac{p^2}{2}\left(\frac{1}{m}-\frac{1}{M}\right)=10.2$

4. Relate this to initial kinetic energy

We want $K = \frac{p^2}{2m}$

From the previous equation: $\frac{p^2}{2m}\left(1-\frac{m}{M}\right)=10.2$

So, $K\left(1-\frac{m}{M}\right)=10.2$

Now, $\frac{m}{M} = \frac{200}{1000} = \frac{1}{5}$

Therefore, $1-\frac{m}{M} = 1-\frac{1}{5}=\frac{4}{5}$

Thus, $K\cdot \frac{4}{5}=10.2$

So, $K=10.2\times \frac{5}{4}=12.75\ \text{eV}$

5. Match with the form $\dfrac{N}{4}$ eV

Given: $K=\frac{N}{4}\ \text{eV}$

But we found: $K=12.75\ \text{eV} = \frac{51}{4}\ \text{eV}$

Hence, $N=51$

6. Comparison with stored answer

Stored correct answer: $51$

Our derived answer is also $51$.

So the stored answer is correct.