Correct option (c) three
Explanation:
Since ionisation potential of hydrogen atom is \(13.6 \mathrm{eV}\).
$$
\therefore \quad E_{1}=-13.6 \mathrm{eV}
$$
Now, \(E_{n}-E_{1}=\frac{-13.6}{n^{2}}-(-13.6)=12.1\)
$$
\begin{aligned}
&\frac{-13.6}{n^{2}}+13.6=12.1 \\
&\therefore \quad n=3
\end{aligned}
$$
After absorbing \(12.1 \mathrm{eV}\), the electron of \(\mathrm{H}\)-atom is excited to \(3^{\mathrm{rd}}\) shell.
Thus, possible transitions we 3 i.e., \(3 \rightarrow 2,2 \rightarrow 1\) and \(3 \rightarrow 1\).
