Question

In a hydrogen atom, an electron of mass $m$ and charge $e$ revolves in an orbit of radius $r$ making $n$ revolutions per second. If the mass of hydrogen nucleus is $M$, the magnetic moment associated with the orbital motion of electron is

• A. $\frac{\pi ne{r}^{2}m}{M+m}$
• B. $\pi ne{r}^2$
• C. $\frac{\pi ne{r}^2}{m}$
• D. $\frac{\pi ne{r}^{2}m}{M}$

Answer 1

The correct option is B $\pi ne{r}^2$

As we know that magnetic moment $\mu$ is given by

$\mu =NiA$

Here, number of turn, $N=1$
and area, $A=\pi r^2$

As charge $e$ revolves in an orbit of radius $r$ making $n$ revolutions per second. So,

Current, $i=\frac{\text{charge}}{\text{Time}}=en$

Thus, magnetic moment $\mu$ is

$\mu =\left(1\right)\left(\pi r^2\right)\left(en\right)$

$\therefore \mu =\pi en{r}^2$

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