Q139 / 150 · MHT CET 2025 (23 Apr Shift 1) · Physics
Let the current 'I' be associated with an electron of charge 'e' moving in a circular orbit of radius ' $r$ ' with speed ' $v$ ' around the positively charged nucleus. The ratio $\frac{r}{v}$ is
- A$\frac{\mathrm{e} \mathrm{I}}{2 \pi}$
- B$\frac{2 \pi}{\mathrm{eI}}$
- Correct answerC$\frac{\mathrm{e}}{2 \pi \mathrm{I}}$
- D$\frac{2 \mathrm{e} \mathrm{I}}{\pi}$
Solution
The current $I$ for an electron of charge $e$ in a circular orbit equals the charge passing a point per unit time. Since charge $e$ passes a point each revolution period $T$, the current is given by $I = \frac{e}{T}$.
The time period for one orbit is $T = \frac{2\pi r}{v}$, where $r$ is the radius and $v$ is the orbital speed.
Combining these yields $I = \frac{e}{\frac{2\pi r}{v}} = \frac{ev}{2\pi r}$.
Rearranging to solve for the ratio $\frac{r}{v}$ gives $\frac{r}{v} = \frac{e}{2\pi I}$.
This expression corresponds to option C.
Hence, the correct answer is option: 'C'.