An electron in a hydrogen atom makes a transition from $$n = 2$$ to $$n = 1$$ and emits a photon. This photon strikes a doubly ionized lithium atom which was already in an excited state and completely removes the orbiting electron. The least quantum number for the excited state of the lithium-ion for the process is

First we calculate the energy of the photon that is emitted when the hydrogen atom makes the transition from the second Bohr orbit ($$n = 2$$) to the first Bohr orbit ($$n = 1$$).

The energy of the $$n^{\text{th}}$$ level in a hydrogen atom is given by the Bohr formula

$$E_n = -\,13.6\ \text{eV}\,\frac{1}{n^2}.$$

So for hydrogen,

$$E_2 = -\,13.6\ \text{eV}\,\frac{1}{2^2} = -\,13.6\ \text{eV}\,\frac{1}{4} = -\,3.40\ \text{eV},$$

and

$$E_1 = -\,13.6\ \text{eV}\,\frac{1}{1^2} = -\,13.6\ \text{eV}.$$

The energy difference between these two levels is the energy of the emitted photon:

$$\Delta E = E_1 - E_2 = \bigl(-\,13.6\ \text{eV}\bigr) - \bigl(-\,3.40\ \text{eV}\bigr) = -\,13.6\ \text{eV} + 3.40\ \text{eV} = -\,10.2\ \text{eV}.$$

The negative sign merely shows that energy is released; the photon energy itself is the magnitude,

$$E_{\text{photon}} = 10.2\ \text{eV}.$$

Now this photon strikes a doubly ionized lithium atom $$\text{Li}^{2+}$$, which is hydrogen-like but with nuclear charge $$Z = 3$$. For any one-electron ion, the Bohr energy formula generalizes to

$$E_n = -\,13.6\ \text{eV}\, \frac{Z^2}{n^2}.$$

Hence for $$\text{Li}^{2+}$$, whose single orbiting electron feels $$Z = 3$$, we have

$$E_n(\text{Li}^{2+}) = -\,13.6\ \text{eV}\,\frac{3^2}{n^2} = -\,13.6\ \text{eV}\,\frac{9}{n^2} = -\,122.4\ \text{eV}\,\frac{1}{n^2}.$$

If the electron is already in the $$n^{\text{th}}$$ excited state, its binding (ionisation) energy—that is, the additional energy needed to remove it completely—is the magnitude of that level energy:

$$E_{\text{bind}} = \Bigl|E_n(\text{Li}^{2+})\Bigr| = \frac{122.4\ \text{eV}}{n^2}.$$

The incoming photon can ionise the electron only if its energy is at least this binding energy, so we must satisfy

$$E_{\text{photon}} \;\ge\; E_{\text{bind}}.$$ $$\Longrightarrow\quad 10.2\ \text{eV} \;\ge\; \frac{122.4\ \text{eV}}{n^2}.$$

Now we solve for $$n^2$$:

$$10.2\,n^2 = 122.4 \quad\Longrightarrow\quad n^2 = \frac{122.4}{10.2} = 12.00.$$

Taking the square root,

$$n = \sqrt{12.00} = 3.464\ldots$$

Because $$n$$ must be an integer quantum number, we choose the next higher integer

$$n_{\text{min}} = 4.$$

This is the least principal quantum number of the excited state of $$\text{Li}^{2+}$$ from which a 10.2 eV photon can just manage to knockout the electron.

Hence, the correct answer is Option B.