I've read from several sources that electromagnetic radiation begins to have an "ionizing" effect right around the time the frequency passes the uv spectrum and into x-ray/gamma ray spectrum. [1] [2] [3]

The reasoning given for this is that the higher frequency waves contain more energy, enough to tear apart molecular bonds.

When I compare this to sound waves it makes sense because high pitched sounds are more damaging to human ears than low pitched sounds are. [4]

However just because a high pitched sound may cause you to go deaf more easily, this doesn't mean my ears would enjoy standing 3 feet away from a 12,000-watt sub-woofer playing a low pitched sound.

In other words, I understand high-frequency waves contain more energy by nature, but if you ramp up the amplitude of the low-frequency waves they can start to do harm too.

So with electromagnetic radiation is there a point that I could say produce infrared waves that would also be ionizing? Or is there something that is inherently different about high-frequency em waves that cause the ionizing effect?


5 Answers 5



Einstein's photoelectric effect theory won him the Nobel prize, and it relates very closely to this. Although it's different from photoionisation, it relies on similar ideas.

His proposition: each atom will absorb the energy of one photon, and the energy of a photon is given by $$E=h\nu$$where $h$ is Plank's constant, a really, really tiny number ($6.62607004 × 10^{-34} m^2 kg / s$), and $\nu$ is frequency of the light. Higher intensity light, which is analogous to wave amplitude, contains more photons, but the energy of each photon is the same for a given frequency.

If I shine low-frequency high-intensity light on a surface, there's plenty of energy, but each atom, upon absorbing one photon, won't be able to lose an electron. However, if I shine high-frequency light, even if the intensity is low, each atom which absorbs a photon will be able to loose an electron, and we see ionization.

But when the intensity is high enough, even low frequencies will cause ionization.

This phenomenon, called multi-photon ionization, occurs when the atom absorbs more than one photon. It's usually pretty rare, because an atom frequently emits the other photons before it absorbs enough energy totally, but at very high enough intensities, it's appreciable.

Sound works differently in air: we generally don't say it's quantized in the same way, although if you examine it more minutely, you'll see that it can be quantized as phonons, which aren't evident in gases. But that's not relevant to your question, it's just something to keep in mind if you want to generalize a bit more by discussing sound in condensed matter.

Oddly enough, the parallel to sound and hearing loss is wrong! See this Biology SE question... high-frequency sounds are dangerous not particularly because they have more energy (which they do, see the equation for sound energy in a container), but because of the nature of the human ear and the alignment of hairs in it.


All frequencies of light can be ionizing. Even a static (zero frequency) field can rip the electrons off atoms if strong enough.

However, when we talk about ionizing radiation, we're usually thinking about quantum effects. In quantum mechanics both sound and light are quantized, into phonons and photons respectively, with energy $$E = \hbar \omega$$ each. And a classical light wave or sound wave is simply a coherent state of many photons/phonons stacked on top of each other.

Light sources with frequency $\omega$ high enough for the energy of a single photon to ionize a single atom are thus especially dangerous, because you can still get hurt even if the total energy of the photons is very low. That is, you can get hurt by X-rays without noticing anything, but you can't get hurt by microwaves without noticing; a single microwave photon has too little energy to do anything. The only way to get hurt by a microwave is for it to heat you up, by interaction with a strong classical microwave field, and there's no way to not notice that.

Sound isn't dangerous in the way that X-rays are. First off, we typically live in air, and a gas like air behaves too messily to really have phonons (see discussion here) at all. Even if air did support phonons, the frequencies the phonons can have are bounded by how fast the air molecules can oscillate to form a sound wave -- more precisely, the wavelength must be much greater than the mean free path. This places an extremely low frequency limit relative to light, so one phonon is always harmless.

While it is true that high-frequency sound is more dangerous, all discussion about the effects of sound on humans is firmly in the classical regime. That is, the danger of high-frequency sound has absolutely nothing to do with the fact that high-frequency phonons have more energy.

  • $\begingroup$ Mechanical oscillators operating at several Mhz while surrounded in air do exist (eg quartz oscillator crystals in appropriate circuits) ... do they actually emit extremely high pitched sound? $\endgroup$ May 26, 2018 at 18:38
  • $\begingroup$ @rackandboneman That’s well below the cutoff for air, so they do emit sound. The frequency of the phonons, notwithstanding the fact that phonons don’t really exist in air, is still much much lower than the threshold for ionization. $\endgroup$
    – knzhou
    May 26, 2018 at 18:43

A multi kW powered CO2 laser can easily strip an atom of many electrons. Such lasers are used to generate extreme ultraviolet (EUV) light for computer chip manufacturing. In such light sources tin atoms are stripped of 10-12 electrons. The subsequent recombination produces EUV light.

  • $\begingroup$ What is the mechanism in this case? Heating of the target and then ionization by collision of the target's atoms with each other? $\endgroup$
    – Ruslan
    May 26, 2018 at 19:13
  • 2
    $\begingroup$ The mechanism is electric dipole transition. Although it happens far off resonance, the enormous field strength ensures a high transition rate. $\endgroup$
    – my2cts
    May 26, 2018 at 19:18

As the other answers have stated, the term "ionizing radiation" has a specific technical meaning, and its use is restricted to radiation where a single photon at the radiation's frequency has enough energy $E=h\nu$ to ionize biological tissue.

However, it is important to note that radiation of all frequencies can produce ionization if its intensity is strong enough. In the literature this is known as multiphoton ionization, and you can think of it heuristically as two or more photons being present at the same time and being absorbed simultaneously.

This does have a price, of course, because the fact that you need those two or more photons to be 'present at the same time' (or, in more technical language, that you're working in perturbation theory of order $\geq 2$) means that the light source needs to be intense enough to allow for that simultaneous presence. This ends up meaning that for an $n$-photon process the ionization rate scales with the light intensity $I$ as $I^n$,* which is a punitively high scaling for any light source that's not a laser.

However, once we did develop lasers, in 1960, it took just five years for the first example of multiphoton ionization to be described, taking the record from single-photon to seven photons all in one go [JETP Lett. 1, 66 (1965)]. In the decades since, multiphoton ionization has been an important tool in the toolbox of atomic and molecular physics for our understanding of the structure and dynamics of matter.

On the other hand, the requirement that the intensity be high does mean that you need specialized experiments to show the effect, which means that ionization by low-frequency radiation is negligible in everyday life $-$ which is why the term 'ionizing radiation' is restricted. In essence, the fact that the ionization rate scales linearly with the intensity means that the total number of ionization events (together with the biological damage it causes if one assumes a linear biological response, which may or may not be a good model (example)) will only depend on the total energy that passes through the tissue, and not the rate at which it is delivered. Thus, if you dilute the dose by turning the intensity down by half but keeping it on for twice as long (so that the total energy in the radiation is constant), then the single-photon ionization will stay constant, but $n$-photon processes will go down by $1/2^n$.

That said, if you want a reasonably accessible demonstration of ionization with a long-wavelength light source, the thing to search for is light-induced optical breakdown in air (example), in which a long-wavelength laser is used to ionize air at its focus. This is mostly an avalanche phenomenon (see this answer for more details) but the seed electrons are often produced via multiphoton ionization.

* for intensities that are low enough for perturbation theory to hold. Most of the interesting physics in multiphoton ionization occurs in non-perturbative regimes where things are more complex. The introduction of my PhD thesis has more details on what that looks like.

  • $\begingroup$ > "the fact that the ionization rate scales linearly with the intensity means that the biological response is only dependent on the radiation dose, and not how it is delivered:" I think this is almost certainly incorrect, although I am not an expert on this topic. I'd desist from making statements on biological responses to ionizing radiation, as it is not a resolved problem and not subject that could be resolved with established physics only. It would be more appropriately discussed on biology/health forums. $\endgroup$ May 30, 2018 at 14:00
  • $\begingroup$ @Ján That's only a statement about the cross-section and it's a part of physics; in fact, the concept of "dose" wouldn't make sense if that statement failed. But yes, I guess that at high enough dose rates the biological response might become nonlinear, and that statement can be sharpened. $\endgroup$ May 30, 2018 at 14:06
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Jun 2, 2018 at 8:38
  • $\begingroup$ I think your edit was an improvement. Still, "linear biological damage" is unclear statement; does it mean the number of cell deaths due to radiation is linear function of dose? If so, it is not implied by the LNT model, because LNT model is not about what happens to individual cells, but it states that radiation increases risk of cancer developing in a lifetime after the exposure linearly with dose. The LNT model is about probability of extreme biological response any time after the exposure, not about the immediate biological response itself. $\endgroup$ Jun 4, 2018 at 9:25
  • $\begingroup$ @Ján "Linear biological damage" means that biological damage (however you wish to define it) scales linearly with the total ionization yield; as marked explicitly, that may or may not be a realistic model and it is for biologists to tell whether it is or not. I see no need for further edits at this time, and the language is plenty clear as it stands. $\endgroup$ Jun 4, 2018 at 9:38

One way in which infrared radiation could be ionizing air would be if the intensity of the radiation was so strong that it would induce electric breakdown of the air, like when static electric field makes a spark. This requires very high electric field strength for static field and for IR radiation, the electric field strength would be probably somewhat less than that, but still high. Perhaps with very strong IR laser this would be possible.

  • $\begingroup$ Well, did you check whether a suitably strong IR laser can actually achieve that effect? Or indeed to research the question at all? $\endgroup$ Jun 4, 2018 at 9:54
  • $\begingroup$ @EmilioPisanty I did not, it wasn't that interesting to me to do research on this. See my2cts's answer for practical details on this effect. $\endgroup$ Jun 5, 2018 at 21:47
  • $\begingroup$ Nope, that's a separate effect - sequential or semi-sequential multiphoton ionization, which can happily run on low-pressure gases that are too rarefied to support avalanches. To see the details of the mechanism you refer to, the answer to refer to is a previous answer of mine, the final link in my answer here. $\endgroup$ Jun 5, 2018 at 23:00
  • $\begingroup$ Ok, that's an interesting distinction. $\endgroup$ Jun 5, 2018 at 23:37

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