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

The angular momentum of an electron in the $n$-th orbit of a hydrogen atom is given by Bohr's quantization rule:

$L = \dfrac{nh}{2\pi}$

Given that the angular momentum is $\dfrac{3h}{\pi}$, we can equate the two expressions:

$\dfrac{nh}{2\pi} = \dfrac{3h}{\pi}$

Solving for $n$, we get:

$n = 6$

The energy of an electron in the $n$-th orbit of a hydrogen atom is given by the formula:

$E_n = -\dfrac{13.6}{n^2} \text{ eV}$

Substituting $n = 6$ into the energy formula:

$E_6 = -\dfrac{13.6}{6^2} \text{ eV}$

$E_6 = -\dfrac{13.6}{36} \text{ eV}$

$E_6 \approx -0.377 \text{ eV} \approx -0.38 \text{ eV}$

Answer: $-0.38$