Question

An electron in the ground state of hydrogen atom is revolving in anti clockwise direction in circular orbit of radius R. The orbital magneitc dipole moment of the electron will be.

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

Answer 1

Answer: (A)

$\frac{eh}{4\pi m}$

Answer 2

According to Boh’s theory,

$mvr = n\frac{h}{2\pi} = \frac{h}{2\pi}$ $(\because n = 1)$

$\therefore v = \frac{h}{2\pi mr}$

We know that rate of flow of charge is current

$\therefore I = e\left( \frac{v}{2\pi r} \right) = \frac{ev}{2\pi r} = \frac{e}{2\pi r} \times \frac{h}{2\pi mr} = \frac{eh}{4\pi r^{2}mr^{2}}$

Magnetic dipole moment = IA

$\therefore M = \frac{eh}{4\pi^{2}mr^{2}} \times \pi r^{2} = \frac{eh}{4\pi m}$