Question:
If $e$ is the electronic charge, $c$ is the speed of light in free space and $\mathrm{h}$ is Planck's constant,
the quantity $\frac{1}{4 \pi \varepsilon_{0}} \frac{|\mathrm{e}|^{2}}{h c}$ has dimensions of :
Correct Option: 1
Solution:
$\mathrm{F}=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{e}^{2}}{\mathrm{r}^{2}}$
$\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$
$\left[\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}} \times \frac{1}{h c}\right]=\frac{\mathrm{Fr}^{2}}{\mathrm{E} \lambda}=\left(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right)$