In any Bohr orbit of the hydrogen atom, the ratio of kinetic energy to potential energy of the electron is
Answer & explanation
Correct answer: option 3
K.E = $\frac{k Z e^2}{2 r}$ and P.E. = $-\frac{k Z e^2}{r}$ ; ∴ $\frac{K . E .}{P . E .}=-\frac{1}{2}$