You are shining ultraviolet light on a gas of an unknown element. You know that

Question

You are shining ultraviolet light on a gas of an unknown element. You know that an electron starts in a ground state with an energy of -14.20 eV. The electron absorbs a 4.00-eV photon. The electron immediately drops to an intermediate energy state Einter of energy -13.00 eV (in this case we will assume it emits a photon when it drops to the intermediate state, though most often the energy is lost through other means). The electron promptly drops back down to the ground state.
What is the energy of the state which the electron is in after absorbing the UV photon?

What is the energy of the first photon emitted?

What is the energy of the second photon emitted and is this process considered fluorescence or phosphorescence? (Both answers must be correct for the answer to be correct.)

fluorescence
phosphorescence

Explanation / Answer

1. It started with E = -14.20 eV, and absorbed 4.00 eV, so that put it in a state of energy
-14.20 eV + 4.00 eV = -10.20 eV

2. The 1st photon was emitted by a drop from -10.20 eV to -13.0 eV, so the energy emitted was
E = -10.20 eV - (-13.0 eV) = 2.80 eV

3. The 2nd photon was emitted by a drop from -13.0 eV to -14.20 eV, so the energy emitted was
E = -13.0 eV - (-14.20 eV) = 1.20 eV

the prompt/immediate emission in such situations, is fluorescence; that in order to be called phosphorescence, there must be a longer delay between absorption and emission, caused by "forbidden" transitions in quantum mechanics.

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