Calculate Energy of Electron in Hydrogen Atom Given Angular Momentum

To find the energy of the electron in the hydrogen atom, we use Bohr's quantization rule for angular momentum and the energy formula for an electron in the $n^{\text{th}}$ orbit.

Step 1: Determine the principal quantum number ($n$) According to Bohr's second postulate, the angular momentum ($L$) of an electron in the $n^{\text{th}}$ orbit is quantized and given by: $L = \frac{nh}{2\pi}$

Given that the angular momentum of the electron is $L = \frac{3h}{\pi}$, we can set the two expressions equal: $\frac{nh}{2\pi} = \frac{3h}{\pi}$

Canceling $h$ and $\pi$ from both sides: $\frac{n}{2} = 3 \implies n = 6$

Step 2: Calculate the energy of the electron in the $n = 6$ orbit The energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by: $E_n = -\frac{13.6}{n^2} \text{ eV}$

Substituting $n = 6$: $E_6 = -\frac{13.6}{6^2} \text{ eV}$ $E_6 = -\frac{13.6}{36} \text{ eV} \approx -0.3778 \text{ eV} \approx -0.38 \text{ eV}$

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

Correct Option: C