Q87 / 120 · VITEEE 2010 · Physics
The electron of hydrogen atom is considered to be revolving round a proton in circular orbit of radius $\mathrm{h}^{2} / \mathrm{me}^{2}$ with velocity $\mathrm{e}^{2} / \mathrm{h}$, where $\mathrm{h}=\hbar / 2 \pi$. The current $i$ is
- A$\frac{4 \pi^{2} \mathrm{me}^{5}}{\hbar^{2}}$
- Correct answerB$\frac{4 \pi^{2} \mathrm{me}^{5}}{\hbar^{3}}$
- C$\frac{4 \pi^{2} \mathrm{~m}^{2} \mathrm{e}^{2}}{\hbar^{3}}$
- D$\frac{4 \pi^{2} \mathrm{~m}^{2} \mathrm{e}^{5}}{\hbar^{3}}$
Solution
Current, $i=\frac{e}{t}=\frac{e}{2 \pi r / v}=\frac{e v}{2 \pi r}$
Here, $v=\frac{e^{2}}{\hbar}$ and $r=\frac{\hbar^{2}}{m e^{2}}$
$$
\begin{array}{l}
\therefore i=\frac{e\left(e^{2} / \hbar\right)}{2 \pi\left(h^{2} / m e^{2}\right)}=\frac{e^{3} \times m e^{2}}{2 \pi h^{3}}=\frac{m e^{5}}{2 \pi h^{3}} \\
\because h=\frac{h}{2 \pi}(\mathrm{given}) \\
\therefore i=\frac{m e^{5}}{2 \pi\left(\frac{\mathrm{h}}{2 \pi}\right)^{3}}=\frac{4 \pi^{2} m e^{5}}{h^{3}}
\end{array}
$$
Hence, the correct answer is option: 'B'.