An electron in the n = 6 level emits a photon with a wavelength of 410.2 nm. To what energy level does the electron move?
A)n = 1
B)n = 2
C)n = 3
D)n = 4
E)n = 5

Question

Quizplus · Accepted Answer

The energy difference between two levels in the hydrogen atom is given by the formula derived from the Rydberg equation: $\Delta E = E_f - E_i = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, where $R_H$ is the Rydberg constant, $n_i$ is the initial energy level, and $n_f$ is the final energy level. The wavelength of the emitted photon is related to the energy difference by the equation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength. Given the wavelength, we can calculate the energy of the emitted photon and then use the Rydberg formula to find the final energy level. The given wavelength corresponds to a transition to the $n = 2$ level for a photon emitted in the Balmer series of the hydrogen spectrum.

An Electron in the N = 6 Level Emits a Photon