Radiation coming from transitions $$n = 2$$ to $$n = 1$$ of hydrogen atoms fall on He$$^{+}$$ ions in $$n = 1$$ and $$n = 2$$ states. The possible transition of helium ions as they absorb energy from the radiation is:

We begin with the well-known Bohr energy formula for any hydrogen-like species:

$$E_n = -\,13.6\,\text{eV}\; \dfrac{Z^{2}}{n^{2}}$$

Here $$Z$$ is the nuclear charge and $$n$$ is the principal quantum number. A photon emitted or absorbed in a transition from level $$n_i$$ to $$n_f$$ carries an energy equal to the magnitude of the difference of these two levels:

$$\Delta E = |E_{n_f}-E_{n_i}|.$$

First we calculate the energy of the radiation coming from the hydrogen transition $$n = 2 \rightarrow n = 1$$. For ordinary hydrogen we have $$Z = 1$$, so

$$E_1 = -13.6\;\text{eV}, \qquad E_2 = -\,13.6\;\dfrac{1}{2^{2}} = -3.4\;\text{eV}.$$

Now, the emitted photon energy is

$$\Delta E_H = |E_1 - E_2| = |-13.6 - (-3.4)| = 10.2\;\text{eV}.$$

This $$10.2\;\text{eV}$$ photon is incident on singly-ionised helium, He$$^{+}$$, for which $$Z = 2$$. Therefore its energy levels obey

$$E_n(\text{He}^{+}) = -13.6\;\dfrac{(2)^2}{n^{2}} = -54.4\;\dfrac{1}{n^{2}}\;\text{eV}.$$

The problem states that some He$$^{+}$$ ions are initially in $$n = 1$$ and some in $$n = 2$$. We must look for every upward (absorptive) transition whose energy difference equals the photon energy $$10.2\;\text{eV}$$.

Case 1: initial level $$n_i = 1$$.

The final level $$n_f$$ must satisfy

$$\Delta E = E_{n_f} - E_{1} = (-54.4/n_f^{2}) - (-54.4) = 54.4\left(\dfrac{1}{1^{2}} - \dfrac{1}{n_f^{2}}\right).$$

Setting this equal to $$10.2\;\text{eV}$$ gives

$$54.4\Bigl(1 - \dfrac{1}{n_f^{2}}\Bigr) = 10.2.$$

Dividing by $$54.4$$,

$$1 - \dfrac{1}{n_f^{2}} = \dfrac{10.2}{54.4} = 0.1875.$$

Hence

$$\dfrac{1}{n_f^{2}} = 1 - 0.1875 = 0.8125,$$ $$n_f^{2} = \dfrac{1}{0.8125} = 1.2308,$$ $$n_f \approx 1.11.$$

This is not an integer greater than 1, so no transition from $$n = 1$$ can absorb the given photon.

Case 2: initial level $$n_i = 2$$.

The energy difference to a higher level $$n_f (>2)$$ is

$$\Delta E = E_{n_f} - E_2 = -\dfrac{54.4}{n_f^{2}} - \Bigl(-\dfrac{54.4}{2^{2}}\Bigr) = 54.4\left(\dfrac{1}{2^{2}} - \dfrac{1}{n_f^{2}}\right).$$

Substituting the photon energy,

$$54.4\left(\dfrac{1}{4} - \dfrac{1}{n_f^{2}}\right) = 10.2.$$

Divide through by $$54.4$$:

$$\dfrac{1}{4} - \dfrac{1}{n_f^{2}} = \dfrac{10.2}{54.4} = 0.1875.$$

Therefore,

$$\dfrac{1}{n_f^{2}} = \dfrac{1}{4} - 0.1875 = 0.25 - 0.1875 = 0.0625.$$

Taking reciprocals,

$$n_f^{2} = \dfrac{1}{0.0625} = 16, \qquad\text{so}\qquad n_f = 4.$$

Thus a $$10.2\;\text{eV}$$ photon can promote a He$$^{+}$$ ion from $$n = 2$$ to $$n = 4$$, but to no other higher level starting from $$n = 1$$ or $$n = 2$$.

The only option that matches this result is

$$n = 2 \;\rightarrow\; n = 4.$$

Hence, the correct answer is Option B.