Atoms & Nuclei

JEE Main level, single correct, medium

30 of 239

Question 30 of 239

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

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Answer: (C) −0.38-0.38

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

L=nh2πL = \dfrac{nh}{2\pi}

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

nh2π=3hπ\dfrac{nh}{2\pi} = \dfrac{3h}{\pi}

Solving for nn, we get:

n=6n = 6

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

En=−13.6n2 eVE_n = -\dfrac{13.6}{n^2} \text{ eV}

Substituting n=6n = 6 into the energy formula:

E6=−13.662 eVE_6 = -\dfrac{13.6}{6^2} \text{ eV}

E6=−13.636 eVE_6 = -\dfrac{13.6}{36} \text{ eV}

E6≈−0.377 eV≈−0.38 eVE_6 \approx -0.377 \text{ eV} \approx -0.38 \text{ eV}

Answer: −0.38-0.38

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