Question

Using Rutherford model of the atom, derive the expression for the total enrgy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?

Answer 1

Let's assume in a hydrogen atom, an electron having a -e charge revolves around the nucleus having a +e charge in a circular orbit of radius r.

Let,

$F_c$ = Centripetal force required by the elctron to move in a circular orbit of radius r.

$F_e$ = Electrostatic force of attraction between the revolving electron and the nucleus.

The electrostatic force of attraction $\left(F_e\right)$ provides the necessary centripetal force,

$F_c=F_e$

$\frac{m{v}^2}{r}=\frac{\left(e\right)\left(e\right)}{4\pi {\epsilon }_0{r}^2}...\left(i\right)$

K.E. of electron in the orbit.

K.E.= $\frac{1}{2}m{v}^2$

From equation (i), K.E.= $\frac{e^2}{8\pi {\epsilon }_{0}r}$

Potential energy of electron in orbit,

P.E.= $\frac{\left(e\right)(-e)}{4\pi {\epsilon }_{0}r}=\frac{-e^2}{4\pi {\epsilon }_{0}r}$

$\therefore$ Total energy of electron in hydrogen atom

$E=K.E.+P.E.=\frac{e^2}{8\pi {\epsilon }_{0}r}-\frac{e^2}{4\pi {\epsilon }_{0}r}$

$E=-\frac{e^2}{8\pi {\epsilon }_{0}r}$

Here, negative sign indicates that the revolving electron is bound to the positive nucleus.