1. Energy of photon emitted by hydrogen for transition $n=2 \to n=1$

For a hydrogen atom, $E_n=-\frac{13.6}{n^2}\text{ eV}$

So, $E_1=-13.6\text{ eV}, \qquad E_2=-3.4\text{ eV}$

Photon energy emitted is $\Delta E = E_2-E_1 = (-3.4)-(-13.6)=10.2\text{ eV}$

So the photon has energy $E_\gamma=10.2\text{ eV}$

2. Energy levels of doubly ionized lithium

Doubly ionized lithium is $\text{Li}^{2+}$, which is a hydrogen-like ion with nuclear charge $Z=3$.

For a hydrogen-like ion, $E_n=-\frac{13.6 Z^2}{n^2}\text{ eV}$

Thus for $\text{Li}^{2+}$, $E_n=-\frac{13.6\times 3^2}{n^2}=-\frac{122.4}{n^2}\text{ eV}$

The ionization energy from the excited state $n$ is the energy needed to remove the electron from that level to infinity: $E_{\text{ion}}=\frac{122.4}{n^2}\text{ eV}$

3. Condition for complete removal of electron

The incident photon completely removes the electron, so its energy must be at least equal to the ionization energy from that excited state.

For the least quantum number $n$, $10.2 \ge \frac{122.4}{n^2}$

So, $n^2 \ge \frac{122.4}{10.2}=12$

Hence, $n \ge \sqrt{12}\approx 3.46$

Therefore, the least integer value is $n=4$

4. Checking options

• A: 2 $E_{\text{ion}}=\frac{122.4}{4}=30.6\text{ eV} > 10.2\text{ eV}$ Not possible.

• B: 3 $E_{\text{ion}}=\frac{122.4}{9}=13.6\text{ eV} > 10.2\text{ eV}$ Not possible.

• C: 4 $E_{\text{ion}}=\frac{122.4}{16}=7.65\text{ eV} < 10.2\text{ eV}$ Possible.

• D: 5 Also possible, but not the least value.

So the correct option is C.

5. Comparison with stored answer

Stored correct answer: C

Derived answer: C

They agree.