Third level of a certain molecule

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To determine the frequency of the photon emitted when a molecule transitions from the third vibrational level to the ground level, the energy difference between these two states must be calculated. The energies provided are 1.15 x 10^-20 J for the ground state and 4.31 x 10^-20 J for the third level. The energy change, ΔE, is found by subtracting the ground state energy from the excited state energy. This energy change corresponds to the photon emitted during the transition. To find the frequency (ν) of the emitted photon, Planck's law is applied, which states that E = hν, where E is the energy of the photon and h is Planck's constant. By rearranging this equation, the frequency can be calculated once ΔE is known.
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The energies of the ground and third level of a certain molecule are 1.15 x 10-20 J and 4.31 x 10-20 J respectively.

What is the frequency (in Hz) of the photon emitted when this molecule drops from the third vibrational level to the ground level?

You don't have to solve it. Could someone just tell me the process and equations?
 
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Hi,
Just apply Planck's law.
 


Hi Westlund,

When you look at a problem it may help to write down the symbols for the things you are given and the things that you are looking for. Then, try and find an equation relating the known and unknowns.

In this case you are given two energies one at ground state (E0) and another at some excited state (Eexcited). In going from an excited to a ground state, there is an accompanying change in energy [the difference between the two energies](\DeltaE). In this case, the energy change is accompanied by an emission of a photon. In order to determine the frequency of this emitted photon (\nu), we must relate it in some way to its energy (\DeltaE). This is where Rajini's comment of using Planck's Law comes in. E = h\nu.

Sorry the "nu's" are not supposed to be superscript.
 
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