JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Modern Physics

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

- 0.04 E_{0} eV

where

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 π L}{h}

is ____ :

(A)2
(B)4
(C)5✓ Correct
(D)6

Answer: (C)

Solution

The angular momentum in Bohr's model is

L = \frac{nh}{2 π}

, so

\frac{2 π L}{h} = n

. The energy of the nth orbit is

E_{n} = - \frac{E _{0}}{n ^{2}}

. Given

E = - 0.04 E_{0} = - \frac{E _{0}}{25}

, we have

n^{2} = 25

, thus

n = 5

. Therefore,

\frac{2 π L}{h} = 5

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option

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