What is the energy of the photon emitted when an electron drops from the $n = 20$ state to the $n = 7$ state in a hydrogen atom?

Question

Quizplus · Accepted Answer

The energy of a photon emitted when an electron drops from a higher energy level to a lower energy level is given by the difference in energy between the two levels. The energy of an electron in a hydrogen atom is given by the equation:$$E = -13.6 \text{ eV} / n^2$$where $n$ is the principal quantum number of the energy level.The energy of the photon emitted when an electron drops from the $n = 20$ state to the $n = 7$ state is:$$E = -13.6 \text{ eV} / 20^2 - (-13.6 \text{ eV} / 7^2) = 0.244 \text{ eV}$$Therefore, the correct answer is A) 0.244 eV.

What Is the Energy of the Photon Emitted When an Electron