What is ratio of Bohr magneton to the nuclear magneton?

m_{p} / m_{e}

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m_{p}^{2} / m_{e}^{2}

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\frac{m_{e}}{m_{p}}

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Solution

The correct option is A $m_{p} / m_{e}$
Bohr magneton is a physical quantity and a unit for expressing the magnetic moment of electron caused by its orbital or spin angular momentum.
$μ_{B} = \frac{e ℏ}{2 m_{e}}$

Similarly nuclear magneton is used to express the same for nucleons.
Thus $μ_{N} = \frac{e ℏ}{2 m_{p}}$

$⟹ \frac{μ_{B}}{μ_{N}} = \frac{m_{p}}{m_{e}}$

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