A hydrogen atom relaxes to a lower energy state by emitting a 2.55 eV photon.What is the lower energy state?
A)n = 1
B)n = 2
C)n = 3
D)n = 4
E)n = 5

Question

Quizplus · Accepted Answer

The energy of a photon emitted by a hydrogen atom during a transition from a higher energy level to a lower energy level is given by the equation E = -13.6 eV/n^2 (where n is the initial energy level). Using this equation and the given energy of the photon, we can solve for n: 
2.55 eV = (-13.6 eV/n^2) - (-13.6 eV/2^2)
2.55 eV = -13.6 eV/n^2 + 3.4 eV
13.55 eV = 13.6 eV/n^2
n^2 = 13.6 eV/13.55 eV
n^2 = 1.003
n = 1.001
Since n must be a positive integer, the closest value is n = 2. Therefore, the hydrogen atom relaxed to a lower energy state with n = 2, and the best choice is B.

A Hydrogen Atom Relaxes to a Lower Energy State by Emitting