Question

An electron in the ground state of hydrogen atom is revolving in anticlockwise direction in circular orbit of radius $R$. The orbital magnetic dipole moment of the electron will be

• A. $\frac {eh}{\pi m}$
• B. $\frac {eh}{4\pi m}$
• C. $\frac {eh^2}{4\pi m}$
• D. $\frac {e^2h}{4\pi m}$

Answer 1

Answer: (B)

According to Bohrs theory,
$mvr = n \frac{h}{2\pi} = \frac{h}{2\pi} \quad (\because n = 1)$
$\therefore v = \frac{h}{2\pi m r}$
We know that rate of flow of charge is current.
$\therefore I = e(\frac{v}{2\pi r}) = \frac{ev}{2\pi r}$
$ = \frac{e}{2\pi r} \times \frac{h}{2\pi m r} $
$ = \frac{eh}{4\pi^2 m r^2}$
Magnetic dipole moment $=I A$
$\therefore M = \frac {eh}{4\pi^2 mr^2} \times \pi r^2$
$=\frac{eh}{4\pi m}$