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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):
A
$\frac{he}{\pi m}$
B
$\frac{he}{2\pi m}$
C
$\frac{heB}{\pi m}$
D
$\frac{heB}{2\pi m}$
Explanation
For $n=1$, flux $\Phi = B \pi r^2 = h/e \implies r^2 = \frac{h}{\pi e B}$. Orbital current is $I = \frac{e^2 B}{2\pi m}$, yielding magnetic moment $\mu = I A = \frac{he}{2\pi m}$.
Detailed Solution
The quantized flux is $\Phi = B A = B (\pi r^2) = n \frac{h}{e}$. For the lowest state ($n = 1$), $r^2 = \frac{h}{\pi e B}$. In a magnetic field, the centripetal balance gives $\frac{m v^2}{r} = e v B \implies v = \frac{e B r}{m}$. The orbital period is $T = \frac{2\pi r}{v} = \frac{2\pi m}{e B}$. The effective current is $I = \frac{e}{T} = \frac{e^2 B}{2\pi m}$. The orbital magnetic moment is $\mu = I A = I (\pi r^2) = \left(\frac{e^2 B}{2\pi m}\right) \left(\frac{h}{e B}\right) = \frac{h e}{2\pi m}$ (Bohr magneton).