What is the energy of the photon emitted when a hydrogen atom undergoes a transition from the $n = 20$ state to the $n = 7$ state?
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 the equation E = hf, where h is Planck's constant and f is the frequency of the photon. The frequency of a photon is given by the equation f = c/\lambda, where c is the speed of light and \lambda is the wavelength of the photon. The wavelength of a photon is given by the equation \lambda = hc/E, where hc is Planck's constant times the speed of light.The energy of the photon emitted when a hydrogen atom undergoes a transition from the $n = 20$ state to the $n = 7$ state is given by the equation:$$E = hf = hc/\lambda = hc/(hc/E) = E$$$$E = (13.6 \text{ eV})[(1/7^2) - (1/20^2)] = 0.244 \text{ eV}$$Therefore, the correct answer is A) 0.244 eV.

What Is the Energy of the Photon Emitted When a Hydrogen