According to the definition on Wikipedia, ionising radiation is radiation which has sufficient energy to remove an electron from an atom. So a high energy gamma ray is definitely ionising, but visible light is not.

What about photoelectric effect? As I understand it, blue light hitting a metal like potassium can free an electron.

So doesn't that make it ionising radiation?

  • $\begingroup$ I think there is some convention to call "ionising" anything from hard UV and above, although your objection is absolutely correct. NB. you can ionize highly excited (i.e. Rydberg) atoms with far infrared or even microwave radiation $\endgroup$
    – dominecf
    Apr 3, 2016 at 19:11
  • $\begingroup$ well, but this would be indeed an ionisation, as opposed to the photoelectric effect, where the atoms are disconnected from the electrons before already. $\endgroup$
    – Ilja
    Apr 4, 2016 at 7:05

1 Answer 1


Well, it's better not to argue about words ... but I would say, that it makes more sense not to consider the photoelectric effect as ionisation. (There is no ion produced anyway, is there? :))

There is nothing special about charges in metals (and there is also nothing special about the slow free electron produced), and nothing "bad", whereas ionising radiation is associated with bad things like cancer etc.

The electron was not tightly bound to an atom anyway, there are plenty of those free electrons in the metal, and you can move them and thansfer them to other pieces of metal with simple electrical fields. The effect has much more in common with heating the metal than with ionisation. You do something to a big reservoire of electrons, not to a single atom.

My conclusion: one can definitely argue which radiation is potentially ionising (see comments to the question), but clearly this very process is not an ionization.

  • $\begingroup$ And depending on the material, the photons may otherwise be classified as ionizing radiation (UV, 5.1 eV for gold), or may be part of the visual spectrum (red-orange, 1.95 eV for cesium) $\endgroup$ Apr 3, 2016 at 19:37

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