Question

A hydrogen atom in the ground state absorbs a 12.75 eV photon. Immediately after the absorption, the atom undergoes a quantum jump to the next-lowest energy level.

What is the wavelength of the photon emitted in this quantum jump?

Express your answer using four significant figures.

I've seen this question before, but I'm looking the wavelength, not the energy, or n. Thanks!

Answer 1

For the hydrogen atom the energy of the quantum state n is E = - (13.6 eV) / n^2. The ground state has n = 1 and E = - 13.6 eV.

The energy of the atom after absorption is -13.6 + 12.75 = -0.85 eV. The quantum number for this energy state is given by

-0.85 = -13.6/n^2
n^2 = 13.6/0.85 = 16
n =4.

The next-lower energy level has energy
E = -13.6/3^2 = -1.511
so the transition emits a photon of energy -0.85-(-1.511) = 0.661 eV = 0.661 * 10^-19

0.661 = hc / lambda

lambda = 6.63x10^-34 Js x 3.00x10^8 m/s? / 0.661 * 10^-19 = 3.02*10^-6 m