Energy of incident beam $=12.5 \mathrm{eV}$
At room temperature, $\mathrm{H}$ will be in ground state whose energy is $\mathrm{E}_1=-13.6 \mathrm{eV}$
$\therefore$ Total energy of electron after absorption will be $\mathrm{E}_1+$ $\mathrm{E}_{\text {incident }}=-13.6+12.5=-1.10 \mathrm{eV}$
This corresponds to a level between $\mathrm{n}=3$ and $\mathrm{n}=4$ (closer to $\mathrm{n}=3$ )
The possibilities for the electron is to jump from $\mathrm{n}^{\text {th }}$ level are
(a) For $\mathrm{n}_2=3,2$ to $_1=1$ within Lyman series
(b) For $\mathrm{n}_2=3$ to $\mathrm{n}_1=2$ in Balmer series.
The wave length for each series would be
(a) For $\mathrm{n}_2=3$ to $\mathrm{n}_1=1$ (Lyman series)
$\mathrm{E}_{31}=\mathrm{E}_3-\mathrm{E}_1=-1.5-(-13.6)=12.1 \mathrm{eV}$
It has wavelength,
$$
\begin{aligned}
\lambda_{31} &=\frac{h \mathrm{c}}{\mathrm{E}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{12.1 \times 1.6 \times 10^{-19}} \\
&=1.02 \times 10^{-7} \mathrm{~m}=102 \mathrm{~nm}
\end{aligned}
$$
For $\mathrm{n}_2=2$ to $\mathrm{n}_1=1$ (Lyman Series)
$\mathrm{E}_{21}=\mathrm{E}_2-\mathrm{E}_1=-3.4-(-13.6)=+10.2 \mathrm{eV}$
$\therefore$ Wavelength emitted,
$$
\lambda_{21}=\frac{h c}{E_{21}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{10.2 \times 1.6 \times 10^{-19}}=126 \mathrm{~nm}
$$
(b) Also for Balmer Series, $\mathrm{n}_2=3, \mathrm{n}_1=2$
$$
\mathrm{E}_{32}=\mathrm{E}_3-\mathrm{E}_2=-1.5-(-3.4)=1.9 \mathrm{eV}
$$
$\therefore$ Wavelength emitted,
$$
\begin{aligned}
\lambda_{32} &=\frac{h c}{\mathrm{E}_{32}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}} \\
&=6.5 \times 10^{-7} \mathrm{~m}=650 \mathrm{~nm}
\end{aligned}
$$
$\therefore$ The excited electron can either jump from $\mathrm{n}=3$ level to $\mathrm{n}=1$ directly giving wavelength $102 \mathrm{~nm}$ or first jump from $n=3$ to $n=2$ level emitting wavelength of $650 \mathrm{~nm}$ and then come to ground state emitting wavelength of $126 \mathrm{~nm}$.