Angular momentum of an electron in a hydrogen atom is $$\frac{3h}{\pi}$$, then the energy of the electron is __________ eV.

For a hydrogen atom, Bohr’s quantisation rule states that the orbital angular momentum $$L$$ of the electron is given by
$$L = n\,\frac{h}{2\pi}$$
where $$n$$ is the principal quantum number.

Here, the angular momentum is given as $$\displaystyle L = \frac{3h}{\pi}$$. Equate this to the quantised expression to find $$n$$:

$$\frac{3h}{\pi}=n\,\frac{h}{2\pi}\quad\Longrightarrow\quad n=\frac{\frac{3h}{\pi}\;(2\pi)}{h}=3\times2=6$$

Thus the electron is in the $$n=6$$ orbit.

The total energy of the electron in the $$n^{\text{th}}$$ Bohr orbit of hydrogen is
$$E_n=-\frac{13.6\text{ eV}}{n^{2}}$$

Substituting $$n=6$$:

$$E_6=-\frac{13.6}{6^{2}}\text{ eV}=-\frac{13.6}{36}\text{ eV}=-0.3777\text{ eV}\approx-0.38\text{ eV}$$

Therefore, the energy of the electron is $$-0.38\text{ eV}$$.

Option C which is: $$-0.38$$