Question

The electron of an H-atom is revolving around the nucleus in circular orbit having radius $\frac{h^2}{4\pi m{e}^2}$ with $\left(\frac{2\pi e^2}{h}\right)$. The current produced due to the motion of electron is

• A. $\frac{2\pi m^2{e}^2}{3h^2}$
• B. zero
• C. $\frac{2{\pi }^{2}me}{h^2}$
• D. $\frac{4{\pi }^{2}m{e}^5}{h^3}$

Answer 1

Answer: (D)

$\frac{4{\pi }^{2}m{e}^5}{h^3}$

In C.G.S. unit the radius of H-atom is: $r=\frac{h^2}{4{\pi }^{2}m{e}^2}$

and the velocity of the electron is: $v=\frac{2\pi e^2}{h}$.

The time period of a revolution is given by: $T=\frac{2\pi r}{v}=\frac{h^3}{4{\pi }^{2}m{e}^4}$.

Thus the current produced during the motion of electron is: $i=e\times \nu =\frac{e}{T}=\frac{4{\pi }^{2}m{e}^5}{h^3}$

where $\nu$ is the frequency of oscillations.