Question

An electron $(e,m)$ is revolving in a circular orbit about the nucleus of an atom. Find a relationship between the orbital magnetic dipole moment $\overrightarrow{\mu }$ and the angular momentum $\overrightarrow{L}$ of the electron about the centre of its orbit.

• A. $\overrightarrow{\mu }=\frac{e}{m}\overrightarrow{L}$
• B. $\overrightarrow{\mu }=-\frac{e}{m}\overrightarrow{L}$
• C. $\overrightarrow{\mu }=-\frac{e}{2m}\overrightarrow{L}$
• D. $\overrightarrow{\mu }=\frac{e}{2m}\overrightarrow{L}$

Answer 1

The correct option is C $\overrightarrow{\mu }=-\frac{e}{2m}\overrightarrow{L}$

Let $v$ be the speed of the electron and $r$ be the radius of its orbit.

The orbiting electron creates an electric current opposite to the velocity.

Thus, current $I$ is charge $e$ times the frequency $f$ of the orbital motion.

$\therefore I=ef=\frac{e}{T}=\frac{ev}{2\pi r}$



The magnitude of magnetic moment,

$\mu =IA=\frac{ev}{2\pi r}\pi r^2=\frac{evr}{2}....\left(1\right)$

The direction of the magnetic dipole moment is perpendicular to the loop.

The orbital angular momentum is given by

${\overrightarrow{L}}_{\text{orbital}}=\overrightarrow{r}\times \overrightarrow{p}$

$|{\overrightarrow{L}}_{\text{orbital}}|=rp\sin {90}^{\circ }=mvr...\left(2\right)$

The direction of the orbital angular momentum is determined from the vector product right-hand rule.

The direction of angular momentum and magnetic dipole moment is opposite to each other. Comparing eqns. $\left(2\right)$ and $\left(1\right)$, we obtain

$\overrightarrow{\mu }=-\frac{e}{2m}\overrightarrow{L}$

Hence, option $\left(c\right)$ is the correct answer.