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 -15.80 eV. The electron absorbs a 4.00-eV photon. The electron immediately drops to an intermediate energy state Einter of energy -14.20 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 first photon emitted?

What is the energy of the second photon emitted and is this process considered fluorescence or phosphorescence?

Explanation / Answer

part a: energy of the state right after absorbing the photon=initial energy + energy of the photon

=-15.8+4=-11.8 eV

part b: as it drop to an intermediate state of energy -14.2 eV, energy of the photon emitted

=energy of initial state-energy of final state=-11.8 eV - (-14.2 eV)

=2.4 eV

part c:

as it finally drops to the ground state, energy of the second photon emitted

=(-14.2 eV)-(15.8 eV)=1.6 eV

as the system does not re-emit the energy instantly and take some time to radiate the acquired energy, it is called Phosphorescence.

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