Question

An electron in the ground state of hydrogen atom in revolving in anticlockwise direction in a circular orbit of radius R (see fig)
If the expression for the orbital magnetic moment of the electron is $M=\frac{eh}{X\pi m}$. Find X?

Answer 1

In ground state $(n=1)$ according to Bohr's theory:
$mvR=\frac{h}{2\pi }$ or $v=\frac{h}{2\pi mR}$

Now time period, $T=\frac{2\pi R}{v}=\frac{2\pi R}{h/2\pi mR}=\frac{4{\pi }^{2}m{R}^2}{h}$

Magnetic moment, $M=iA$
Where, $I=\frac{charge}{\text{time period}}=\frac{e}{\frac{4{\pi }^{2}m{R}^2}{h}}=\frac{eh}{4{\pi }^{2}m{R}^2}$
and $A=\pi R^2$
$\therefore M=\left(\pi R^2\right)\left(\frac{eh}{4{\pi }^{2}m{R}^2}\right)$ or $M=\frac{eh}{4\pi m}$
Direction of magnetic moment $\overrightarrow{M}$ is perpendicular to the plane of orbit.