Question

The ratio of the magnetic dipole moment to the angular momentum of the electron in the $1^{st}$ orbit of hydrogen atom is

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

Answer 1

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

Magnetic dipole moment $\mu =IA$

$\therefore$ $\mu =e\nu \left(\pi r^2\right)=e\times \frac{v}{2\pi r}\times \pi r^2=\frac{evr}{2}$ .......(1)

Angular momentum $L=mvr$

$\therefore$ Ratio $\frac{\mu }{L}\frac{evr/2}{mvr}=\frac{e}{2m}$