The energy of the photon emitted when an electron jumps from n = 4 to n = 2 is given by ΔE = 13.6 [1/2² - 1/4²] eV. What is ΔE?
Q1. The energy of the photon emitted when an electron jumps from n = 4 to n = 2 is given by ΔE = 13.6 [1/2² - 1/4²] eV. What is ΔE?
Answer: 2.55 eV
Explanation: ΔE = 13.6 [1/4 - 1/16] = 13.6 * (3/16) = 2.55 eV.