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If the frequency of light is $f$ and if $f \ge f_t$, where $f_t$ is the threshold frequency, electrons are emitted if light is shined on a metal surface. By my understanding, the light comes in and is absorbed by the atom. The atom then has too much energy and emits an electron. I don't remember much about nuclear decay, but when an atom has too much energy it can emit an alpha particle or a gamma ray or whatever else (something about decaying down to a stable line).

Anyways, so if an atom has too much energy and ejects an electron, is that considered a nuclear decay? Also, does the photoelectric effect then ionize materials? Could I take a metal sheet, shine light on it, and get a metal sheet that is now positive? Could I do this at home?

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    $\begingroup$ Nuclear decay - root word "nucleus". Is the photoelectric effect a type of nuclear decay : No. $\endgroup$ – AlanZ2223 Sep 6 '14 at 19:50
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    $\begingroup$ No, but is there an equivalent of the photoelectric effect for nuclei? Sure. If you ramp up the energy of a gamma into the MeV range or higher, it can knock constituents out of nuclei in non-elastic scattering events, just as visible or near UV light can liberate electrons from atoms or metals. Can you do the photoelectric effect at home? Sure. It's not a completely trivial experiment, but it can be done. Are you asking for instructions? $\endgroup$ – CuriousOne Sep 6 '14 at 20:00
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The photoelectric effect involves the shell (electrons) of the atom. Photon energies are typically in the range of electron volts.

To excite the nucleus, you'll need higher energies (MeV or hundreds of keV).

What you are probably looking for is called photonuclear reaction.

Anyways, so if an atom has too much energy and ejects an electron, is that considered a nuclear decay?

If this electron is coming from the shell, then it's NOT a decay of the nucleus (aka nuclear decay). If however, this is coming from a $\beta^-$ decay (bound neutron decays to a bound proton plus an electron and an anti-electron neutrino) of the nucleus, then yes, this is a nuclear decay.

One has to be careful when talking about 'the atom has too much energy': one has to clarify whether it's an electron in the shell which is not in the energetically lowest possible state or whether it's the nucleus which is not in its energetically lowest state.

Could I take a metal sheet, shine light on it, and get a metal sheet that is now positive?

One would have to apply some force, e.g. by means of an electric field, to get the electrons to move away from the sheet of metal. Also, one would have to do this in vacuum in order to avoid the electrons being slowed down by the gas molecules. A phototube for example is a device which works in this way.

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    $\begingroup$ Not induced gamma emission. The photoelectric effect involves the emission of a particle that was previously a part of the atom. The nuclear equivalent is reactions like ${}^{Z}X + \gamma \to {}^{Z-1}X + n$ or ${}^{Z}_{A}X + \gamma \to {}^{Z-1}_{A-1}Y + p$. There are many examples, but I can't be bothered to look any up. $\endgroup$ – dmckee Sep 6 '14 at 20:35
  • $\begingroup$ The article on induced gamma emission I linked starts with the sentence: "In physics, induced gamma emission (IGE) refers to the process of fluorescent emission of gamma rays from excited nuclei". That's what the original poster was looking for as an analogy to the photoelectric effect, isn't it ? $\endgroup$ – Andre Holzner Sep 6 '14 at 21:19
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    $\begingroup$ No, the photoelectric effect is incident photons causing the emission of electrons. So the best nuclear analogy is incident photons causing the emission of nucleons. You've got something unspecified causing the emission of photons which is time reversed from what you want. $\endgroup$ – dmckee Sep 6 '14 at 22:28
  • $\begingroup$ good point, so 'photonuclear reaction' is probably more adequate. $\endgroup$ – Andre Holzner Sep 7 '14 at 7:17
  • $\begingroup$ Yeah. That's a good word for the situation. $\endgroup$ – dmckee Sep 7 '14 at 15:54

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