A light of energy 12 . 75   eV is incident on a hydrogen atom in its ground state. The atom absorbs the…

Let the electron jumped to $n^{th}$ excited state.

In the ground state, energy $E=-13.6 \mathrm{eV}$

So, using relation $12.75=13.6\left[\frac{1}{1^2}-\frac{1}{n^2}\right]$

$$\begin{gathered} \Rightarrow 0.9375=\left[1-\frac{1}{n^2}\right] \\ \Rightarrow n^2=16 \end{gathered}$$

$\Rightarrow n=\sqrt{16}=4$

Now, angular momentum, $L=\frac{nh}{2\pi }=\frac{4h}{2\pi }=\frac{2h}{\pi }$

 $=\frac{2}{\pi }\times 4.14\times {10}^{-15}$

$=\frac{828\times {10}^{-17}}{\pi } \mathrm{eVs}$

Hence, the value of $x=828$.