The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is

1. 1 : 1

2. 2 : $-$1

3. 1 : $-$1

4. 1 : $-$2

Doubtly Asked 30 Oct 2026 NEET UG 2018 Practice test neet Save WhatsApp Report

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Answer: 3. 1 : $-$1

In a Bohr orbit, kinetic energy $K=\dfrac{ke^2}{2r}$ and potential energy $U=-\dfrac{ke^2}{r}$.

Total energy $E=K+U=-\dfrac{ke^2}{2r}=-K$.

$$\frac KE=\frac{K}{-K}=1:-1$$

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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