Question

In any Bohr orbit of the hydrogen atom, the ratio of kinetic energy to potential energy of the electron is .

• A. 1/2
• B. 2
• C. $- 1/2$
• D. – 2

Answer 1

Answer: (C)

$- 1/2$

Answer 2

$K.E = \frac{kZe^{2}}{2r}$ and $P.E. = - \frac{kZe^{2}}{r};$ ∴ $\frac{K.E.}{P.E.} = - \frac{1}{2}$.