The energy of an electron in an orbit of the Bohr's atom is -0.04… — Solution | JEE Physics

MCQEasyJEE 2026Bohr's Model & Hydrogen Spectrum

The energy of an electron in an orbit of the Bohr's atom is $-0.04 E_g$ where $E_g$ 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 $\frac{2 \pi L}{h}$ is _____ :

A
$4$
B
$6$
C
$5$
D
$2$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The energy of the electron is $E_n = -0.04 E_g$ .

Find: The value of $\frac{2 \pi L}{h}$ .

According to the Bohr model, the energy in the $n^{\text{th}}$ orbit varies as $\frac{1}{n^2}$ and the angular momentum is quantized as

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

Using the given energy,

E_n = -0.04 E_g = -\frac{1}{25} E_g

Comparing with the Bohr energy relation,

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

we get

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

Hence,

n = 5

Now from

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

we obtain

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

Therefore, the value of $\frac{2 \pi L}{h}$ is $5$ . The correct option is C.

Common mistakes

Using the energy relation incorrectly as proportional to $\frac{1}{n}$ instead of $\frac{1}{n^2}$ . This gives a wrong value of $n$ . Always use the Bohr relation that orbital energy varies as $\frac{1}{n^2}$ .
Ignoring the sign and fractional form of $-0.04 E_g$ . Here $0.04 = \frac{1}{25}$ , so the comparison must be made carefully. Convert the decimal to a fraction before solving for $n$ .
Not recognizing that $\frac{2 \pi L}{h} = n$ directly from Bohr quantization. After finding $n$ , substitute it into $L = \frac{n h}{2 \pi}$ instead of trying an unrelated formula for angular momentum.

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