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(B) The magnetic moment $M$ of an electron revolving in the $n$-th orbit is given by the formula: $M = \frac{n e h}{4 \pi m}$.From this expression, we

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$2 \mu_{B}$

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(B) The magnetic moment $M$ of an electron revolving in the $n$-th orbit is given by the formula: $M = \frac{n e h}{4 \pi m}$.From this expression, we can see that the magnetic moment is directly proportional to the principal quantum number $n$, i.e., $M \propto n$.For the first orbit $(n = 1)$, the magnetic moment is $M_1 = \mu_{B}$.For the second orbit $(n = 2)$, the magnetic moment is $M_2 = 2 	imes M_1$.Therefore, $M_2 = 2 \mu_{B}$.

The magnetic moment of an orbital electron in the first orbit of an $\text{H}$-atom is $\mu_{B}$. What is the magnetic moment of that electron in the second orbit?

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