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: \(-0.38\)