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When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon. How does the change in energy level compare to the energy of the emitted photon?

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  • $\begingroup$ Welcome to chemistry.SE. I edited your question to make the title and tags actually relevant. Please feel free to take the tour to find out more about the site. In future try to make your titles actually useful as this is the first thing people see when they see your question so titles like 'chemistry periotic table' don't really say anything about the question's content. $\endgroup$ – bon May 13 '15 at 18:48
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When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon.

That is one of several pathways for deexcitation. Often, internal conversion, the radiationless deeexcitation happens instead.

How does the change in energy level compare to the energy of the emitted photon?

It depends on the relaxed geometry of the excited state, as compared to the ground state.

Remember that the excitation happens on a femtosecond scale. That is amazingly fast, light travels a distance of just $\mathrm{0.3\,\mu m}$ in $\mathrm{1\,fs}$.

Since it takes longer than some femtoseconds to move the much heavier nuclei, the molecule in its electronically excited state still has the geometry of the ground state. This can be far from the lowest vibrational level (the relaxed geometry) of the excited state.

Before emission of a photon occurs, ectronically excited molecules relax and adopt their lowest vibrational state.

This means that the energy of the emitted photon often is (much) lower than the excitation energy. This effect is known as Stokes shift.

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I wanted to add a couple of figures to elaborate on Klaus's answer. An electron "lives" in quantized energy levels, but the atom also vibrates in quantised energy levels, called vibrational states. There are many vibrational states within each main quantum number $n$ (K shell, L shell, etc.). These can be "crudely" approximated by the harmonic oscillator, or more accurately by Morse potentials. Morse potentials are depicted in the figures as the curved lines, one for each main quantum number (e.g. $n=1$ and $n=2$). The horizontal lines within the potential are the various vibrational states. Note that the first vibrational state is not at the bottom of the potential; even at 0 K there will be at least some vibrations!

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Figure 1. Morse potentials for two main quantum numbers. Source: Atkins' "Physical Chemistry", 9th ed, page 505.

The absorption of radiation is depicted by an arrow going from the lowest vibrational state in the lowest electronic state, up to a high vibrational state in the next electronic state. The tiny arrows illustrate the "vibrational relaxation" explained by Klaus. This is a non-radiative process. When the electron reaches the lowest possible vibrational state in $n=2$, then an electronic transition takes place. Note that the electron does not necessarily de-excite directly to the lowest vibrational level. Similarly, note that during the absorption, the electron is not just excited to the lowest vibrational state of $n=2$, but some higher vibrational state. So we have such vibrational relaxation in both $n=1$ and $n=2$. The energy lost during vibrational relaxation leads to the emitted photon being of lower energy than the incident photon. But why is the electron excited to a high vibrational state, and not just to the lowest possible?

The transition in the figure is a vibronic transition, meaning that both a vibrational and an electronic transition occur. Also note that the two Morse potentials in both figures are not perfectly aligned above each other; the upper one is shifted to the right. Now look at Figure 2. The colored curves are the probability distributions of where to find the electron in each vibrational state. See how the number of nodes in each vibrational state increase by 1 for each vibrational transition? The most likely electron position shifts toward the edges for each vibrational state. At the point of convergence of the vibrational state, the electron vibrates so much that it is ejected out of its state and becomes free.

Back to why the electron does not simply jump to the lowest vibrational state in $n=2$. The Franck-Condon principle states that the most likely vibrational state in $n=2$ for the electron, is the one with which the current vibrational state overlaps with the most. The two ground state vibrational states does not overlap significantly, and so such a transition will be very weak. However, as is depicted by the blue arrow in Figure 2, a transition from the ground vibrational state in $n=1$ overlaps very well with the third vibrational ($\nu ' = 2$) state in $n=2$, and so this transition will be very strong. Finding itself $\nu ' = 2$, the electron relaxes to $\nu ' = 0$, and so undergoes de-excitation to $n=1$ ($E_0$ in the figure). This time the best overlap is also for the third vibrational state. Then the electron relaxes some more. Note that several vibronic transitions occur, but some are much more probable than others.

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Figure 2. Probability densities for each vibrational state. Source: Source

So that is why the emitted wavelength is longer than the excitation wavelength; it is due to vibrational relaxation occurring due to the vibronic transition. The Franck-Condon principle gives a way to predict the vibrational transition based on the vibrational wave-function overlaps in the initial and final state.

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  • $\begingroup$ But for the green transition in Figure 2, the energy of the emitted x-ray has exactly the difference in energy between the two orbitals. That is less than the energy of the excitation x-ray. $\endgroup$ – MaxW Oct 23 '15 at 7:40
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When an electron de-excites from a higher energy orbital to a lower energy orbital it loses energy by emitting a photon. Since energy must be conserved, the energy of this photon is equal to the amount of energy the electron has lost, or in other words the difference in energy of the two orbitals involved. The frequency of the photon can then be calculated using $E=hf$ where $h$ is Planck's constant and has a value of $\mathrm{6.626\times10^{-34}~J~s}$.

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