Question

The ratio of magnetic dipole moment of an electron of charge 'e' and mass 'm' in Bohr's orbit in hydrogen atom to its angular momentum is.

• A. $\frac{e}{m}$
• B. $\frac{m}{e}$
• C. $\frac{2m}{e}$
• D. $\frac{e}{2m}$

Answer 1

The correct option is D $\frac{e}{2m}$

Let the velocity with which an electron moves in a circular orbit of radius $r$ be $v$.

Current flowing $I=e\nu =\frac{ev}{2\pi r}$

Area of the circular orbit $A=\pi r^2$

$\therefore$ Magnetic moment of the electron $\mu =IA=\frac{evr}{2}$

Angular momentum $L=mvr$

Ratio of magnetic dipole moment to angular momentum $\frac{\mu }{L}=\frac{evr/2}{mrv}=\frac{e}{2m}$