The frequency of a spectral line is given by f = ΔE / h, where ΔE is the energy difference between two states. What is the frequency of the photon emitted when an electron jumps from n = 3 to n = 2?
Q1. The frequency of a spectral line is given by f = ΔE / h, where ΔE is the energy difference between two states. What is the frequency of the photon emitted when an electron jumps from n = 3 to n = 2?
Answer: 4.57 * 10^14 Hz
Explanation: First, find ΔE = 1.89 eV = 1.89 * 1.6 * 10^-19 J, then use f = ΔE / h to get f = 4.57 * 10^14 Hz.