Question

For separation $r$ between gold nucleus and an $\alpha$-particle, the magnitude of repulsive force is given by

• A. $F=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{2e^2}{r^2}$
• B. $F=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{e^2}{r^2}$
• C. $F=\frac{158\cdot e^2}{4\pi {\epsilon }_0{r}^2}$
• D. $F=\frac{56e^2}{4\pi {\epsilon }_0{r}^2}$

Answer 1

Answer: (C)

For separation $r$, force between $\alpha$-particle (charge $2e$ ) and gold nucleus (charge $Ze$ ) is
$F=\frac{1}{4\pi {\epsilon }_0}\times \frac{2e(Ze)}{r^2}$
For gold,
$Z=79$
$\therefore F=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{158e^2}{r^2}$