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 -18.60 eV. The electron absorbs a 4.00-eV photon. The electron immediately drops to an intermediate energy state Einter of energy -17.80 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). After several seconds the electron 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 second photon emitted and is this process considered fluorescence or phosphorescence? (Both answers must be correct for the answer to be correct.)

Explanation / Answer

the ground state energy is -18.60 eV. since the electron absorbs 4.00 eV, so, its new state with energy is,

E=E0+4.00 eV

=-18.60 eV+4.00 eV

=-14.6 eV.

first teh electron comes to -17.80 eV state and gives first photon. and then it comes to -18.60 eV from -17.80 eV which gives second photon. therefore, the energy of second photon must be,

E= -17.80 eV-(- 18.60 eV)

=0.8 eV

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