If a hydrogen atom,originally in its ground state of energy -13.6 eV,absorbs a photon of energy 15.0 eV,what is the resulting kinetic energy of the electron if the proton has negligible kinetic energy?
A)Such a photon cannot be absorbed in this case.
B)-1.4 eV
C)1.4 eV
D)15.0 eV

Question

Quizplus · Accepted Answer

When a hydrogen atom absorbs a photon, the electron can move from its ground state to an excited state. If the photon has enough energy, the electron can be excited all the way to the ionization potential, at which point it will escape the nucleus completely.

In this case, the photon has energy 15.0 eV, which is greater than the energy of the ground state (-13.6 eV), so the electron can be excited to a higher energy level. The energy of the excited state will be:

E_excited = E_ground + E_photon 
          = (-13.6 eV) + (15.0 eV) 
          = 1.4 eV

Because the proton has negligible kinetic energy, all of the energy from the excited electron will be in the form of kinetic energy. Therefore, the resulting kinetic energy of the electron is 1.4 eV.

If a Hydrogen Atom,originally in Its Ground State of Energy