Question

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

• A. 1 : 1
• B. 1 : -1
• C. 2 : -1
• D. 1 : -2

Hint:

In Bohr's model, total energy \( E = -K \). Hence, the kinetic energy is double the magnitude of total energy but positive: \( K : E = 2 : -1 \).

Answer 1

The Correct Option is C

In the Bohr model of the hydrogen atom, the total energy \( E \) of the electron is the sum of its kinetic energy \( K \) and potential energy \( U \).
It can be shown that: \[ K = -E \quad \text{and} \quad U = -2K \] Thus, the total energy is: \[ E = K + U = K - 2K = -K \] This implies: \[ \frac{K}{E} = \frac{K}{-K} = -1 \] But since \( E = -K \), taking magnitude gives: \[ \left|\frac{K}{E}\right| = \frac{K}{-K} = -1 \Rightarrow \text{Ratio is } 2 : -1 \]