A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron is $n(h/e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ($m$ is the mass of the electron)
Answer: (B) $\dfrac{he}{2\pi m}$
Lowest state, $n = 1$: $B\pi r^2 = \dfrac{h}{e}$, so $r^2 = \dfrac{h}{\pi eB}$.
In the magnetic field, $r = \dfrac{mv}{eB}$, so $v = \dfrac{eBr}{m}$.
Magnetic moment of a circulating charge:
$$\mu = IA = \frac{ev}{2\pi r}\cdot\pi r^2 = \frac{evr}{2} = \frac{e^2Br^2}{2m}$$
Substituting $r^2$:
$$\mu = \frac{e^2B}{2m}\cdot\frac{h}{\pi eB} = \frac{he}{2\pi m}$$
Solution by Sreeraj P, M.Sc Physics