Q. If the electron of a hydrogen atom is present in the first orbit, the total energy of the electron is
A) $\frac{- e^{2}}{r}$
B) $\frac{- e^{2}}{r^{2}}$
C) $\frac{- e^{2}}{2r}$
D) $\frac{- e^{2}}{2r^{2}}$
Explanation
(c)
Kinetic energy =$\frac{1}{2}mv^{2},$
Potential energy $= \frac{{- e}^{2}}{r}$
But, $mv^{2} = \frac{e^{2}}{r}$
KE=$\frac{1}{2}\frac{e^{2}}{r}$
Total energy =KE+PE
$= \frac{1}{2}\frac{e^{2}}{r} - \frac{e^{2}}{r} = \frac{e^{2}}{r}\left( \frac{1}{2} - 1 \right) = \frac{- e^{2}}{2r}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.