Question

The KE of the electron in an orbit of radius $r$ in hydrogen atom is (e = electronic charge)

• A. $\frac{e^2}{r}$
• B. $\frac{e^2}{2r}$
• C. $\frac{e^2}{4r}$
• D. $\frac{e^2}{2r^2}$

Answer 1

The correct option is B $\frac{e^2}{2r}$

The centrifugal force acting on the revolving electron is balanced by the electrostatic force acting between the electron and the nucleus of hydrogen atom so that the electron revolves in a certain fixed orbit.

$\therefore$ $F_{centrifugal}=F_{electrostatic}$

OR $\frac{m{v}^2}{r}=\frac{k{e}^2}{r^2}$

We get $m{v}^2=\frac{k{e}^2}{r}$

Thus kinetic energy of electron $K.E=\frac{1}{2}m{v}^2=\frac{k{e}^2}{2r}=\frac{e^2}{2r}$ (putting $k=1$ in some unit)