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Suppose we have an atom with several energy levels (e.g. an hydrogen), and it is hit by photons.

I know that in order to have the atom change energy levels, the photon must have an energy level precisely equal to the energy difference of the two levels.

What happens if the photon has more energy than the ionization energy for the ground state (i.e., the sum of the series of differences between consecutive levels)?

It is not clear to me. Thanks!

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    $\begingroup$ It is not precisely true that the photon must have exactly the right amount of energy to lift an electron to a higher energy level (though this is a good place to start from in understanding the physics). There are processes in which the photon scatters from the atom promoting an electron as it does so, and short-lived states not exactly on an allowed energy level (these are important in multi-photon processes). $\endgroup$
    – dmckee --- ex-moderator kitten
    Apr 3, 2011 at 18:03

3 Answers 3

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If the photon has more energy than the ionization energy, it can (and often does) ionize the atom.

One nice way to think about this is to put the various energy levels of the atom in an energy level diagram like this:

enter image description here

(taken from this web site). All of the bound-state energy levels lie below the line marked $E_\infty=0$ eV. Above this line is the region called "free electrons" in the diagram. This corresponds to an ionized atom. In this range, the energy levels are not quantized -- that is, any energy is possible. So any photon with enough energy to put the total energy somewhere in that shaded region is capable of ionizing the atom.

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    $\begingroup$ So will the photon be absorbed completely and the electron will have the excessive energy (as kinetic energy)? $\endgroup$
    – R S
    Apr 3, 2011 at 17:46
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    $\begingroup$ @R S: Yes. Or with enough energy the photon can scatter (i.e. you get a lower energy photon in the final state) and still leave the electron free and possessed of some kinetic energy. $\endgroup$
    – dmckee --- ex-moderator kitten
    Apr 3, 2011 at 18:04
  • $\begingroup$ When high energy photons hit lead barriers for example, they keep loosing their energy with every collision untill they have sufficiently low energy to fly right trough the barrier. $\endgroup$
    – MaDrung
    Jun 14, 2017 at 5:59
  • $\begingroup$ @Ted Bunn Why the energy levels above $\infty$ don't have index? Aren't these states also eigenstates of the Hamiltonian, i.e. solution to Schrodinger's equation? $\endgroup$
    – Antonios Sarikas
    Jan 3, 2022 at 20:42
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The photoelectric effect (http://en.wikipedia.org/wiki/Photoelectric_effect) is a phenomenon whereby the surface atoms are ionised through liberation of some of their electrons. These electrons can then leave with a kinetic energy equal to the incident photon energy minus the 'work function' which was the energy required to ionise the atom.

The work function is the energy required to remove the electron from the surface of a material.

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  • $\begingroup$ You should rethink "the electrons are ionised" $\endgroup$
    – Georg
    Apr 3, 2011 at 20:32
  • $\begingroup$ fair point, that's sloppy terminology $\endgroup$
    – Nic
    Apr 3, 2011 at 20:54
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If your photon is sufficiently energetic, in the final state you can find any configuration allowed with the energy-momentum conservations laws, i.e., an excited atom and a lower-energy photon(s): $\gamma (\omega) + A_0 \rightarrow \gamma '(\omega') + A_n,$

$ \omega'<\omega$.

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