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?
A) 0.244 eV
B) 0.264 eV
C) 0.283 eV
D) 0.303 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 the n = 20 state to the n = 7 state in a hydrogen atom is given by the Rydberg formula:$$f = \frac{1}{h} \left( \frac{1}{\lambda} ight) = \frac{R_H}{hc} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$$where R_H is the Rydberg constant, h is Planck's constant, c is the speed of light, n_f is the final energy level, and n_i is the initial energy level.Plugging in the values for n_f, n_i, R_H, h, and c, we get:$$f = \frac{1}{h} \left( \frac{1}{\lambda} ight) = \frac{R_H}{hc} \left( \frac{1}{7^2} - \frac{1}{20^2} ight) = 2.46 	imes 10^{14} Hz$$The energy of the photon is then:$$E = hf = (6.63 	imes 10^{-34} J \cdot s)(2.46 	imes 10^{14} Hz) = 0.244 eV$$Therefore, the correct answer is A.

What Is the Energy of the Photon Emitted When an Electron