Identify the species

The ion is $\mathrm{Li^{++}}$, which is a hydrogen-like ion with: $Z = 3$

It has only one electron, so Bohr's formula for hydrogen-like atoms applies.

Energy of the $n^{\text{th}}$ Bohr orbit

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

For $\mathrm{Li^{++}}$, substitute $Z=3$: $E_n = -\frac{13.6\times 3^2}{n^2} = -\frac{122.4}{n^2}\,\text{eV}$

Find energies of the first and third orbits

• For $n=1$: $E_1 = -122.4\,\text{eV}$

• For $n=3$: $E_3 = -\frac{122.4}{9} = -13.6\,\text{eV}$

Energy required for excitation from $n=1$ to $n=3$

Required energy is the difference: $\Delta E = E_3 - E_1$ $\Delta E = (-13.6) - (-122.4)$ $\Delta E = 108.8\,\text{eV}$

Match with the options

$108.8\,\text{eV}$ corresponds to Option B.

Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

Hence, they agree.