Question

Let's analyze the 'planetary' motion of the electron around the nucleus for a hydrogen atom. Say the electron (mass m_e) is in orbit around the nucleus at a distance of r. How will this orbital radius r be related to the electron's speed v?

• A. $r=\frac{e}{\sqrt{4\pi {ϵ}_0}}.\frac{1}{v}$
• B. $r=\frac{4\pi {ϵ}_0{m}_e}{e^2}.v^2$
• C. $r=\frac{e^2}{4\pi {ϵ}_0{m}_e}.v^2$
• D. $r=\frac{e^2}{4\pi {ϵ}_0{m}_e}.\frac{1}{v^2}$

Answer 1

The correct option is D $r=\frac{e^2}{4\pi {ϵ}_0{m}_e}.\frac{1}{v^2}$

If electron, of mass m_e, were to go around the nucleus with a speed v at a distance r, the electrostatic force of attraction of the nucleus will have to provide the centripetal acceleration required for orbital motion. Applying Newton's second law, we can write -

Centripetal force= $\frac{m_e{v}^2}{r}$= Electrostatic force of attraction = $\frac{z{e}^2}{4\pi {ϵ}_0{r}^2}$
Cancelling r from both denominators and rearranging,

$r=\frac{e^2}{4\pi {ϵ}_0{m}_e{v}^2}$. (z=1 for Hydrogen)