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Atoms and Nuclei Part 3

Section: Physics

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Question : 23 of 98

Marks: +1, -0

The energy of an electron in an orbit of the Bohr's atom is

-0.04 E_0 \text{ eV}

E_0

is the ground state energy. If

L

is the angular momentum of the electron in this orbit and

h

is the Planck's constant, then

\; \frac{2 \pi L}{h}

\_\_\_\_

[21 Jan 2026 Shift 2]

6
2
5
4

Solution:

Concept:

In Bohr's model, the energy of an electron in the

n^{\text{th}}

orbit is

E_n = \frac{E_0}{n^2}

, where

E_0

is the ground state energy.

Angular momentum is quantized as

L = n \frac{h}{2\pi}

.

Explanation:

Given

E_n = -0.04 \, E_0

eV.

Using the energy formula:

\frac{E_0}{n^2} = -0.04 \, E_0

.

Cancel

E_0

and ignore sign (only magnitude matters):

\frac{1}{n^2} = 0.04

.

Thus

n^2 = \frac{1}{0.04} = 25

, so

n = 5

.

Angular momentum

L = n \frac{h}{2\pi} = 5 \frac{h}{2\pi}

.

Therefore

\frac{2\pi L}{h} = \frac{2\pi}{h} \times 5 \frac{h}{2\pi} = 5

.

Answer:

\boxed{5}

, which corresponds to option C.

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