Energy levels of hydrogen

For hydrogen atom, $E_n=-\frac{13.6}{n^2}\,\text{eV}$

Initially, the atom is in ground state $n=1$, so $E_1=-13.6\,\text{eV}$

Energy absorbed

Incident light has energy $12.75\,\text{eV}$

If the atom absorbs this photon and goes to level $n$, then $E_n-E_1=12.75$

Substitute the hydrogen energy formula: $-\frac{13.6}{n^2}-(-13.6)=12.75$ $13.6\left(1-\frac{1}{n^2}\right)=12.75$

Solve for $n$

$1-\frac{1}{n^2}=\frac{12.75}{13.6}$

Now, $\frac{12.75}{13.6}=0.9375=\frac{15}{16}$

Hence, $1-\frac{1}{n^2}=\frac{15}{16}$ $\frac{1}{n^2}=\frac{1}{16}$ $n^2=16 \Rightarrow n=4$

So the atom reaches the excited state with principal quantum number $n=4$.

Angular momentum in Bohr model

In the $n$th orbit, angular momentum is $L_n=\frac{nh}{2\pi}$

For $n=4$, $L=\frac{4h}{2\pi}=\frac{2h}{\pi}$

Given $h=4.14\times10^{-15}\,\text{eV s}$

Therefore, $L=\frac{2\times4.14\times10^{-15}}{\pi}$ $L=\frac{8.28\times10^{-15}}{\pi}\,\text{eV s}$

Match with given form

The question gives angular momentum as $\frac{x}{\pi}\times10^{-17}\,\text{eV s}$

So, $\frac{x}{\pi}\times10^{-17}=\frac{8.28\times10^{-15}}{\pi}$

Multiply both sides by $\pi$: $x\times10^{-17}=8.28\times10^{-15}$

Hence, $x=8.28\times10^2=828$

Final answer

$\boxed{828}$

Comparison with stored answer

Stored correct answer = $828$

This matches the derived answer.