According to the Rutherford nuclear model of the atom which is a electrical neutral sphere of a very small, massive and positively charged nucleus at the centre surrounding by the revolving electron in their respective stable orbits. The electrostatic force of attraction, $\mathrm{F}_{e}$ between the revolving electrons and the nucleus provides the requisite centripetal force to keep them in their orbits.

For a stable orbit in a hydrogen atom,

$\mathrm{F}_{e}=\mathrm{F}_{c} \quad \text { where } \mathrm{F}_{e}=\text { Electric force }$

$\mathrm{F}_{c} =\text { Centripetal force }$

$\therefore \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{e^{2}}{r^{2}}=\frac{m v^{2}}{r}$

$\therefore r=\frac{e^{2}}{4 \pi \epsilon_{0} \cdot m v^{2}} \ldots \text { (1) }$

which is the relation between the velocity of electron to the orbital radius. The kinetic energy of electron in hydrogen atom,

$\frac{1}{2} m v^{2}=\frac{e^{2}}{2 \times 4 \pi \epsilon_{0} r} \quad[\because \text { From equation (1)] }$

$\therefore K=\frac{e^{2}}{8 \pi \epsilon_{0} r}$$….(2)$

And potential energy,

$U=-\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Z e \times e}{r}$

$\therefore U=-\frac{e^{2}}{4 \pi \epsilon_{0} r}$$…(3)$ $[\because$ For hydrogen $Z=1]$

In this formula, the negative sign in $U$ signifies that the electrostatic force is in the $-r$. Hence, the total energy of electron in a hydrogen atom is, $\mathrm{E}=\mathrm{K}+\mathrm{U}$

$=\frac{e^{2}}{8 \pi \epsilon_{0} r}-\frac{e^{2}}{4 \pi \epsilon_{0} r} \quad[\because$ From equation $(1)$ and $(2)$]