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(B) The magnetic moment $M$ of an electron moving in a circular orbit of radius $R$ with speed $v$ is given by $M = iA$,where $i$ is the equivalent cu

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$n$

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(B) The magnetic moment $M$ of an electron moving in a circular orbit of radius $R$ with speed $v$ is given by $M = iA$,where $i$ is the equivalent current and $A$ is the area of the orbit.The current $i$ is defined as the charge $e$ passing per unit time $T$,so $i = \frac{e}{T} = \frac{e \omega}{2 \pi} = \frac{ev}{2 \pi R}$.The area of the orbit is $A = \pi R^2$.Thus,the magnetic moment is $M = iA = \left( \frac{ev}{2 \pi R} \right) (\pi R^2) = \frac{evR}{2}$.Multiplying and dividing by the mass of the electron $m$,we get $M = \frac{e(mvr)}{2m} = \frac{eL}{2m}$,where $L = mvr$ is the orbital angular momentum of the electron.According to Bohr's quantization postulate,the angular momentum $L$ of an electron in the $n$th orbit is $L = \frac{nh}{2 \pi}$.Substituting this into the expression for $M$,we get $M = \frac{e}{2m} \left( \frac{nh}{2 \pi} \right) = \left( \frac{eh}{4 \pi m} \right) n$.Since $e$,$h$,and $m$ are constants,it follows that $M \propto n$.Therefore,the correct option is $B$.

Magnetic moment due to the motion of the electron in the $n$th energy state of a hydrogen atom is proportional to ..........

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