The angular momentum of an electron moving in the circular path of diameter D in uniform magnetic field $B_0$ is

$(e\, B_0\, D^2)/4$

The correct answer is Option (4) → $(e\, B_0\, D^2)/4$

For an electron of charge $e$ and mass $m$, moving in a uniform magnetic field $B_0$:

The magnetic force provides the centripetal force:

$\frac{m v^2}{r} = e v B_0$

$\Rightarrow v = \frac{e B_0 r}{m}$

Angular momentum:

$L = m v r = m \cdot \frac{e B_0 r}{m} \cdot r = e B_0 r^2$

Since $r = \frac{D}{2}$,

$L = e B_0 \left(\frac{D}{2}\right)^2 = \frac{e B_0 D^2}{4}$

Therefore, $L = \frac{e B_0 D^2}{4}$