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 $$\frac{N}{4}$$. The value of $$N$$ is: (Given the mass of the hydrogen atom to be $$1\,\text{GeV c}^{-2}$$)........

We have a particle whose rest-mass is given as $$m = 200\,\text{MeV}\,c^{-2}$$. The hydrogen atom, initially at rest, has rest-mass $$M = 1\,\text{GeV}\,c^{-2} = 1000\,\text{MeV}\,c^{-2}$$. The particle strikes the atom, and immediately after the collision the following facts are stated:

1. The particle itself is brought to rest, so its final momentum is zero.
2. The hydrogen atom recoils (so it acquires kinetic energy) and is simultaneously promoted to its first excited state.

The first excited state of hydrogen corresponds to the principal quantum number $$n = 2$$. Using the well-known Bohr energy formula, the ground-state energy is $$E_1 = -13.6\,\text{eV}$$, while for $$n = 2$$ we have

$$E_2 = \frac{E_1}{n^{2}} = \frac{-13.6}{4}\,\text{eV} = -3.4\,\text{eV}.$$

Hence the excitation (internal) energy absorbed by the atom is

$$\Delta E = E_2 - E_1 = (-3.4) - (-13.6)\,\text{eV} = 10.2\,\text{eV}.$$

We now apply the two conservation laws.

Conservation of linear momentum
Initially only the incoming particle has momentum. If its initial momentum is $$p$$, then after the collision the same momentum must be carried by the recoiling hydrogen atom because the particle is at rest. Therefore

$$p_{\text{initial}} = p_{\text{final}} \quad\Longrightarrow\quad p = p_{\text{H}},$$

where $$p_{\text{H}}$$ is the recoil momentum of the hydrogen atom.

Classical (non-relativistic) expression for kinetic energy
Because the kinetic energies we shall obtain turn out to be of order only a few electronvolts, they are completely negligible compared with the rest-energies of hundreds of MeV. Hence it is consistent to use the non-relativistic formula

$$K = \frac{p^{2}}{2m}.$$

Let $$K$$ be the particle’s initial kinetic energy. Then

$$K = \frac{p^{2}}{2m} \quad\Longrightarrow\quad p^{2} = 2mK.$$

The hydrogen atom’s recoil kinetic energy $$K_{\text{H}}$$ is obtained from the same momentum $$p$$:

$$K_{\text{H}} = \frac{p^{2}}{2M}.$$

Substituting $$p^{2} = 2mK$$ into this expression we get

$$K_{\text{H}} = \frac{2mK}{2M} = \frac{m}{M}\,K.$$

Conservation of total energy
All of the particle’s initial kinetic energy is converted into (i) the recoil kinetic energy of the hydrogen atom and (ii) the internal excitation energy $$\Delta E = 10.2\,\text{eV}$$. Therefore

$$K = K_{\text{H}} + \Delta E.$$

Replacing $$K_{\text{H}}$$ by $$\dfrac{m}{M}K$$ gives

$$K = \frac{m}{M}K + \Delta E.$$

Now we isolate $$K$$ algebraically. First, move the term $$\dfrac{m}{M}K$$ to the left:

$$K - \frac{m}{M}K = \Delta E.$$

Factorising $$K$$ yields

$$K\left(1 - \frac{m}{M}\right) = \Delta E.$$

Hence

$$K = \frac{\Delta E}{1 - \dfrac{m}{M}}.$$

We substitute the numerical values. The mass ratio is

$$\frac{m}{M} = \frac{200\,\text{MeV}\,c^{-2}}{1000\,\text{MeV}\,c^{-2}} = 0.2.$$

Therefore

$$1 - \frac{m}{M} = 1 - 0.2 = 0.8.$$

So the required kinetic energy of the incident particle is

$$K = \frac{10.2\,\text{eV}}{0.8} = 12.75\,\text{eV}.$$

The problem statement writes this energy in the form $$\dfrac{N}{4}\,\text{eV}$$. Setting

$$\frac{N}{4} = 12.75 \quad\Longrightarrow\quad N = 12.75 \times 4 = 51.$$

Hence, the correct answer is Option 51.