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

Answer: (D)

Angular momentum,
$L=m_{e} V r \,\,\,\,...(i)$
Therefore, orbital motion of electron is equivalent to a current,
$I=e(1 / T)$
Period of revolution of electron is given by
$T=\frac{2 \pi r}{v} $
$ \therefore \,\,\,I=e\left(\frac{1}{2 \pi r / . v}\right)=\frac{e v}{2 \pi r}$
Area of electron orbit, $A=\pi r^{2}$
Magnetic dipole moment of the atom
$M=\frac{e V}{2 \pi r} \times \pi r^{2}=e v r$
Using Eq. (i), we have
$M=\left(\frac{e}{2 m_{e}}\right) L$