Question

In any Bohr orbit of hydrogen atom, the ratio of kinetic energy to potential energy of revolving electron at a distance \( r \) from the nucleus is

• A. \( -1 \)
• B. \( +\frac{1}{2} \)
• C. \( 1 \)
• D. \( -\frac{1}{2} \)

Hint:

In Bohr model, potential energy is always twice the magnitude of kinetic energy but negative.

Answer 1

The Correct Option is D

Step 1: Write expressions for energies.
For electron in Bohr orbit: \[ \text{K.E.} = \frac{1}{2} m v^2 \] \[ \text{P.E.} = -\frac{ke^2}{r} \]
Step 2: Use centripetal force condition.
\[ \frac{mv^2}{r} = \frac{ke^2}{r^2} \Rightarrow mv^2 = \frac{ke^2}{r} \]
Step 3: Find ratio.
\[ \frac{\text{K.E.}}{\text{P.E.}} = \frac{\frac{1}{2}mv^2}{-\frac{ke^2}{r}} = -\frac{1}{2} \]
Step 4: Conclusion.
The ratio is \( -\frac{1}{2} \).