# Expected observation if one assumes that light is a wave instead of a particle in the photoelectric effect

For the purpose of this question I am pretending that I have not conducted a photoelectric effect experiment yet and I don't know what the outcome of the experiment will be. My questions are about the expected consequences if one assumes that light is a wave and not a particle (which we all know was not the result of this experiment).

1. A sinusoidal wave is non-localised. A metal target that absorbs a light wave should absorb that energy over a non-zero time interval since the wave is spread out over a non-zero length of space. This suggests that you should be able to liberate electrons from a target if you shine a low-energy wavelength of light onto a target (with energy significantly below the work function) by simply waiting long enough until enough energy is absorbed by the target.

2. Often it is stated that increasing the intensity of a low-energy wavelength light is sufficient to allow photoelectrons to be liberated from the metal if light is a wave.

My questions are:

Is my reasoning for 1. correct that you would expect that shining a low-energy color of light should allow electrons to be liberated from a metal if one simply waits long enough until enough energy is absorbed by the metal?

I don't understand the reasoning behind 2. Increasing the intensity of a source of light increases the number of light waves from your source (the lamp in the experiment), but shouldn't one can expect that each of these light waves will strike a different spot on the metal. Therefore increasing the intensity should liberate more electrons due to more spots on the metal receiving light, but I don't see why increasing the intensity should immediately cause photoelectrons to become liberated (unless all the additional light waves happen to strike the exact same spot on the metal when the intensity is increased).

• Why don't you read fro example the wikipedia article on the photoelectric effect and then ask a more focussed question? Or is this homework? – my2cts Apr 21 at 20:07
• Not homework. I'm trying to figure out an intuitive way to present this experiment, and I want to say, "If light is a wave, we expect X, and if light is a particle we expect Y". Wikipedia seems to confirm the first point. I have seen in various other sources the second point, but this doesn't make sense to me. A more focussed question for point 2 might be stated as 'Will increasing the intensity liberate more electrons, but not liberate them any more rapidly than you would from a less intense source?' – Bob Apr 21 at 20:17
• I think you have a point with this question. Indeed, what does classical electromagnetism predict. – my2cts Apr 21 at 22:54

By assuming light is a wave and not a particle, are you assuming classical physics is correct? There are no photons?

If so, your point 1) is right. That was one of the mysteries of the photoelectric effect that was hard to explain classically. A light of low intensity absorbed by a metal should add energy to the metal slowly. Eventually there should be enough to eject electrons.

However, what was observed was the electrons were never ejected for low frequency light, or they were ejected within a nanosecond or so for higher frequency. Thinking classically, this mean that the energy that was deposited uniformly over the surface of the metal was somehow concentrated very quickly onto a single electron. It was difficult to imagine a process that would do this.

For 2), again thinking classically, a wave deposits energy in proportion to its intensity. If the wave is spread over the surface and is low intensity, energy should continuously be added to the metal. You might expect the energy to stay spread out. If so, the energy near any particular atom increases slowly. But eventually it does increase to a level that might eject an electron.

Of course, things did not turn out as classical thoughts would lead one to expect.

• Thanks. It does seem like in the classical picture we would expect that electrons would be liberated very fast since the speed of light is very fast (and energy would accumulate rapidly). In principle we should be able to increase this timescale by shining a very (very, very, very) long wavelength of light on the target. – Bob Apr 21 at 23:02
• With regard to the second point, wouldn't a higher intensity of light imply more light waves, which again would each (presumably) strike a different electron in the metal? Saying that the wave is "spread over the surface" in the classical picture of light seems to imply that there was only one single light wave being emitted from the lamp rather than many individual light waves that each deposit the same energy? – Bob Apr 21 at 23:05
• Think of classical ocean waves. More energetic waves means bigger waves, not more waves. – mmesser314 Apr 23 at 23:37
• So you are saying that for classical electromagnetic waves a very bright lamp will not create many waves filling a volume, just a single wave of very large amplitude? – Bob Apr 25 at 16:28
• Let us assume there is a filament of surface area A in a lightbulb. I can measure the intensity of the light being emitted by that filament. If I were to cut filament in half I would measure half the intensity of light. Each small area (dA) of the filament's surface emits electromagnetic waves, independent of whether the rest of the filament is there or not. I agree that all the waves from each dA of the filament will interfere and all the properties that come with const/destr interference, but it still seems more correct to think of a room being filled with many waves rather than 1 big wave. – Bob Apr 25 at 16:33

We could make an analogy with sound waves to understand what is the classical outcome. If we are hit by a very loud noise, we can damage our ears, no mater it has low (like an explosion) or high pitch.

But being exposed to a low intensity sound for a long time is not dangerous. It is what we experience in our normal life.

So I think that the picture described in (2) is the correct one for a classical description.

• I don't follow the analogy. A high intensity source of light would consist of many individual light waves. I don't know how well this would translate to acoustics sound is a compression wave (with the wave itself implicitly spread out over an area perpendicular to the direction of travel). – Bob Apr 22 at 2:13