The energy difference between two states is given by ΔE = hf. What is the frequency of the photon emitted when an electron jumps from n = 4 to n = 3?
Q1. The energy difference between two states is given by ΔE = hf. What is the frequency of the photon emitted when an electron jumps from n = 4 to n = 3?
Answer: 1.6 * 10^14 Hz
Explanation: First, find ΔE = E4 - E3 = -0.85 - (-1.51) = 0.66 eV, then use f = ΔE / h to get f = 1.6 * 10^14 Hz.