Some hydrogen like atoms in ground state absorb 'n' photons having same energy and on de-excitation it emits exactly 'n' photons. Then the energy of absorbed photon may be

.Number of dark lines(in absorption) i.e excitation = number of bright lines(in emission), i.e de-excitation
It is possible only when the e - is excited to $n=2$ from ground state
Clearly,  $\mathrm{\Delta E}=91.8\mathrm{eV}\text{ and }40.8\mathrm{eV}$ are possible
The energy of electron in the first and second orbit is  $-13.6\mathrm{eV}\text{ and }-3.4\mathrm{eV}$.
The energy difference is $13.6-3.4=10.2\mathrm{eV}$

a $\frac{40.8}{10.2}=4$ an integer

b $\frac{91.8}{10.2}=9$.an integer

$\begin{array}{l}n=1\Rightarrow 2\to 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\Rightarrow 10.2Z^2\\ 10.2Z^2=91.8\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\Rightarrow Z=3\\ 10.2Z^2=40.8\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\Rightarrow Z=2\end{array}$