Question

Electron of charge \( e \) moves in hydrogen atom (radius \( r \)). Coulomb force between nucleus and electron is:
(Here \( K = \frac{1}{4 \pi \varepsilon_0} \))

• A. \( -K \cdot \frac{e^2}{r^3} \hat{r} \)
• B. \( K \cdot \frac{e^2}{r^3} \hat{r} \)
• C. \( -K \cdot \frac{e^2}{r^2} \hat{r} \)
• D. \( K \cdot \frac{e^2}{r^2} \hat{r} \)

Hint:

Remember Coulomb’s Law: attractive force = negative sign; use \( K = \frac{1}{4\pi \varepsilon_0} \)

Answer 1

The Correct Option is C

Coulomb force: \( \vec{F} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{e^2}{r^2} \hat{r} = K \cdot \frac{e^2}{r^2} \hat{r} \)
Direction is attractive \( \Rightarrow \vec{F} = -K \cdot \frac{e^2}{r^2} \hat{r} \)