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

When a hydrogen atom in the ground state absorbs a 13.32-eV photon, the electron gets promoted to the excited state, n=2. The energy of the photon is equal to the energy difference between the ground state and the excited state. 
Now, we know that the ionization energy of hydrogen is 13.6 eV. So, if the electron were to be promoted to n=1, it would have reached the ionization energy level and would have been completely removed from the atom. Therefore, the electron must have been promoted to the next available energy level, which is n=3. 
The energy difference between n=2 and n=3 is:

E(3)-E(2) = -13.6 eV(1/3^2 - 1/2^2) = -1.51 eV

Therefore, the photon with energy 13.32 eV is sufficient to promote the electron from n=1 to n=3, and the answer is choice C.

A Hydrogen Atom in the Ground State Absorbs a 13