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 -19.10 eV. The electron absorbs a 4.00-eV photon. The electron immediately drops to an intermediate energy state Einter of energy -17.10 eV. 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 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
To determine the energy of the state which the electron is in after absorbing the UV photon, we need to calculate the change in energy between the ground state and the intermediate state.
Given that the ground state has an energy of -19.10 eV and the intermediate state has an energy of -17.10 eV, we can subtract the two values to find the change in energy:
Change in energy = (Energy of intermediate state) - (Energy of ground state)
= -17.10 eV - (-19.10 eV)
= -17.10 eV + 19.10 eV
= 2.00 eV
Therefore, the energy of the state which the electron is in after absorbing the UV photon is 2.00 eV.
To determine the energy of the first photon emitted when the electron drops back down to the ground state, we need to calculate the change in energy between the intermediate state and the ground state.
Change in energy = (Energy of ground state) - (Energy of intermediate state)
= -19.10 eV - (-17.10 eV)
= -19.10 eV + 17.10 eV
= -2.00 eV
Therefore, the energy of the first photon emitted is 2.00 eV.
To determine the energy of the second photon emitted and whether this process is fluorescence or phosphorescence, we need additional information. Fluorescence involves the emission of light immediately after absorption, while phosphorescence involves delayed emission of light.
Since the electron drops back down to the ground state "after several seconds," it suggests a delayed emission, which is characteristic of phosphorescence.
The energy of the second photon emitted would be the same as the energy change between the intermediate state and the ground state, which we calculated above as -2.00 eV.
Therefore, the energy of the second photon emitted is -2.00 eV, and the process is considered phosphorescence.
To determine the energy of the state which the electron is in after absorbing the UV photon, we can subtract the energy of the ground state (initial state) from the energy of the intermediate energy state (final state).
Efinal = Einter - Einitial
Efinal = -17.10 eV - (-19.10 eV)
Efinal = -17.10 eV + 19.10 eV
Efinal = 2.00 eV
Therefore, the energy of the state which the electron is in after absorbing the UV photon is 2.00 eV.
To determine the energy of the first photon emitted, we can use the same logic. The energy of the first photon emitted is equal to the energy of the final state after absorbing the UV photon.
Efirst photon emitted = Efinal = 2.00 eV
Now, to determine the energy of the second photon emitted, we need to know the energy difference between the intermediate energy state (Einter) and the ground state (Einitial).
Esecond photon emitted = Einter - Einitial
Esecond photon emitted = -17.10 eV - (-19.10 eV)
Esecond photon emitted = -17.10 eV + 19.10 eV
Esecond photon emitted = 2.00 eV
Therefore, the energy of the second photon emitted is 2.00 eV.
Since the electron drops back down to the ground state after emitting the second photon, this process is considered fluorescence.