The frequency of revolution of the electron in the nth orbit of hydrogen atom is given by f = me⁴ / (4ε0²n³h³). What is f for n = 2?
Q1. The frequency of revolution of the electron in the nth orbit of hydrogen atom is given by f = me⁴ / (4ε0²n³h³). What is f for n = 2?
Answer: 8.2 * 10^14 Hz
Explanation: Substitute n = 2 in the given formula to get f = me⁴ / (4ε0²2³h³) = 8.2 * 10^14 Hz.