A gas of hydrogen-like ions is prepared in such a way that ions are only in the ground state and the first excited state. A monochromatic light of wavelength 1218 $\text{Å}$ is absorbed by the ions. The ions are lifted to higher excited states and emit radiation of six wavelength, some higher and some lower than the incident wavelength. Calculate the values of the minimum wavelengths (in $\text{Å}$).

$\text{ Incident photon energy }E=\frac{12431}{1218}=10.2\mathrm{eV}$

When six wavelengths are emitted that means electrons are excited to n = 4 and by absorption of 10.2 eV if electrons excite from n₁ to n₂ = 4 we use

$10.2=13.6Z^2\left(\frac{1}{n_1^2}-\frac{1}{16}\right)$

For Z = 1 this is not possible
For Z = 2 we get n₁=2
For Z = 3 onward this is not Possible
Thus Z = 2 and transition is from n₁ = 2 to n₂ : 4
Minimum wavelength (maximum energy) emission is for
transition 4 $\to$ 1

$\Rightarrow {\lambda }_{min}=\frac{12431}{12.75\times 4}=243.74\text{Å}\approx 244\text{Å}$