1 Bohr magneton is

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$\frac{e h}{4 π m_{e}}$

$\frac{e h}{π m_{e}}$

$\frac{e h}{2 π m_{e}}$

$\frac{4 π m_{e}}{e h}$

answer is C.

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Detailed Solution

Magnetic moment of electron in ground state electron of hydrogen atom is known as Bohr magneton. $\frac{M}{L} = \frac{q}{2 m} \Rightarrow M = (\frac{h}{2 π}) (\frac{e}{2 m})$

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