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JEE Main 2013 (9 Apr, Shift 1) · Bohr Model: Orbits, Speed & Angular Momentum

Bohr Model: Orbits, Speed & Angular Momentum question from JEE Main 2013 (9 Apr, Shift 1)

Q114JEE Main 2013 · 9 APR · SHIFT 1Medium

Bohr Model: Orbits, Speed & Angular Momentum

In the Bohr model an electron moves in a circular orbit around the proton. Considering the orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in $n^{th}$ excited state, is:

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Answer: (c)

$M = IA = \frac{ev}{2\pi r}\cdot\pi r^2 = \frac{e}{2m}(mvr)$
Bohr: $mvr = \frac{nh}{2\pi}$
$M = \left(\frac{e}{2m}\right)\frac{nh}{2\pi}$ , so the answer is (c).

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