A hydrogen atom in the ground state absorbs a 13.32-eV photon.To what level is the electron promoted? (The ionization energy of hydrogen is 13.6-eV).

Question

Quizplus · Accepted Answer

The energy required to promote an electron from the ground state to the n=7 level is given by the difference in energy between the energy levels:
ΔE = Efinal − Einitial = −RH/n²final + RH/n²initial
where RH is the Rydberg constant (2.18 × 10⁻¹⁸ J), and n is the principal quantum number.

In this case, Einitial = −13.6 eV (ground state energy of hydrogen), and Efinal = −13.6 eV + 13.32 eV = −0.28 eV (energy of the electron after absorbing the 13.32 eV photon).

Solving for nfinal yields:

−0.28 eV = −RH/n²final + 13.6 eV

n²final = RH/(13.6 eV − 0.28 eV)

nfinal = 7

So the electron is promoted to the n=7 level. Therefore, choice C is the correct answer.

A Hydrogen Atom in the Ground State Absorbs a 13