What is the magnitude of the energy difference between the lowest energy of a hydrogen atom (n = 1) and the second excited state (n = 3)?
A)27.4 eV
B)9.07 eV
C)1.51 eV
D)109 eV
E)12.1 eV

Question

Quizplus · Accepted Answer

The energy levels of a hydrogen atom are given by the formula E_n = -13.6 eV/n^2. For n=1, E_1 = -13.6 eV, and for n=3, E_3 = -13.6 eV/9 = -1.51 eV. The energy difference is |E_1 - E_3| = |-13.6 eV + 1.51 eV| = 12.1 eV.

What Is the Magnitude of the Energy Difference Between the Lowest