Identify the ion and use hydrogen-like atom formula

The ion is $\mathrm{Li^{2+}}$ (written as $\mathrm{Li^{++}}$), which is a hydrogen-like ion with atomic number $Z=3.$

For a hydrogen-like species, the energy of the $n^{\text{th}}$ Bohr orbit is $E_n=-\frac{13.6Z^2}{n^2}\,\text{eV}.$

So for $\mathrm{Li^{2+}}$, $E_n=-\frac{13.6\times 9}{n^2}=-\frac{122.4}{n^2}\,\text{eV}.$

Use the number of spectral lines to find the excited level

If the electron is excited to level $n$, then on de-excitation to the ground state, the total number of possible spectral lines is $N=\frac{n(n-1)}{2}.$

Given that a total of 6 spectral lines are observed, $\frac{n(n-1)}{2}=6$ $n(n-1)=12$ $n^2-n-12=0$ $(n-4)(n+3)=0$

Hence, $n=4.$

So the electron was excited from $n=1$ to $n=4$.

Find the excitation energy

Energy required to excite from $n=1$ to $n=4$ is $\Delta E=E_4-E_1.$

Now, $E_1=-122.4\,\text{eV}, \qquad E_4=-\frac{122.4}{16}=-7.65\,\text{eV}.$

Therefore, $\Delta E=(-7.65)-(-122.4)=114.75\,\text{eV}.$

Convert this energy into wavelength

The absorbed photon energy is $\Delta E=\frac{hc}{\lambda}.$

Using the standard relation $\lambda(\text{nm})=\frac{1240}{E(\text{eV})},$ we get $\lambda=\frac{1240}{114.75}\,\text{nm}.$

$\lambda\approx 10.8\,\text{nm}.$

Match with the options

$\boxed{\lambda=10.8\,\text{nm}}$

So the correct option is A.

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

Hence, they agree.