An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The colomb force $\overrightarrow{\mathrm{F}}$
between the two is

Given data: Radius = r 

Concept Used: Coulomb’s law 

Detailed Solution:

We solve it by coulomb’s law, the force between two charges is,  

$\overrightarrow{\mathrm{F}}=\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^3}\overrightarrow{\mathrm{r}}\text{ where }\mathrm{r}=\text{ distance between charges. }$

$\text{ Here, }{q}_1=-\mathrm{e}\text{, charge of an electron };{\mathrm{q}}_2=\mathrm{Ze}=\mathrm{e}\text{, charge of the nucleus }$

For hydrogen atom, atomic number = 1 = Z

Thus, the force between electron and nucleus will be, $\mathrm{F}=\mathrm{k}\frac{(-\mathrm{e})\mathrm{e}}{{\mathrm{r}}^3}\overrightarrow{\mathrm{r}}=-\mathrm{k}\frac{{\mathrm{e}}^2}{{\mathrm{r}}^3}\overrightarrow{\mathrm{r}}$