Find the energy of the photon emitted when an electron drops from the n = 20 state to the n = 7 state in a hydrogen atom.
A)0.244 eV
B)0.263 eV
C)0.283 eV
D)0.302 eV

Question

Quizplus · Accepted Answer

The energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency of the photon. The frequency of a photon emitted when an electron drops from a higher energy level to a lower energy level is given by f = (E_i - E_f)/h, where E_i is the initial energy of the electron and E_f is the final energy of the electron. In this case, the electron drops from the n = 20 state to the n = 7 state, so the energy of the photon emitted is given by:E = hf = (E_20 - E_7)/h = (-13.6 eV/n^2) * (1/20^2 - 1/7^2) = 0.244 eVTherefore, the correct answer is A.

Find the Energy of the Photon Emitted When an Electron