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JEE Main 2026 (21 Jan, Shift 2) · Bohr Model: Orbits, Speed & Angular Momentum

Bohr Model: Orbits, Speed & Angular Momentum question from JEE Main 2026 (21 Jan, Shift 2)

Q14JEE Main 2026 · 21 JAN · SHIFT 2Medium

Bohr Model: Orbits, Speed & Angular Momentum

The energy of an electron in an orbit of the Bohr's atom is $-0.04E_0$ 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

$\frac{2\pi L}{h}$ is ________:

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Answer: (c)

$E_n = \frac{E_0}{n^2}$ : $\frac{1}{n^2} = 0.04 \Rightarrow n = 5$ .
$L = \frac{nh}{2\pi} \Rightarrow \frac{2\pi L}{h} = n = 5$
So the answer is (c) 5.

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