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Bohr's Model of Atom

The ratio of ...

Question

The ratio of magnetic dipole moment of an electron of charge e and mass m in the Bohr orbit in hydrogen to the angular momentum of the electron in the orbit is:

A

e/m

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B

e/2m

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C

m/e

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D

2m/e

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Solution

The correct option is B e/2m
Magnetic moment $μ = I A = (e ν) A$
$∴$ $μ = \frac{e v}{2 π R} \times π R^{2} = \frac{e v R}{2}$
Angular momentum $L = m v R$
Thus $\frac{μ}{L} = \frac{\frac{e v R}{2}}{m v R} = \frac{e}{2 m}$

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