Did you read
ZapperZ's post? Firstly, be aware that the situation can be a lot more complicated. Needless to say, to be "absolutely correct" (no such thing by the way, you can only be "more correct"), you would have to think a lot harder than what I shall present below. But at the level that I have been discussing this issue, a few intuitive ideas could suffice in constructing a fairly consistent view. Let me try again.
monish said:
What doesn't happen? Don't you liberate more electrons the longer you shine the light? and doesn't higher intensity make this happen faster?
I think these things do happen. According to you, isn't that consistent ("one would immediately expect") with the wave model?
What doesn't happen:
when the incoming light is of frequency lower than the threshold, then it doesn't matter how long you shine the light nor what is the intensity, no photoelectrons are emitted (this is an experimental observation)
what would a wave-model predict:
firstly, the argument I used last time was based on the idea of "energy transfer". I thought you would understand what that means. Anyway, a (classical) wave-model would predict that even when frequency is below the threshold, if you shine light on the metal plate for long enough, electrons should be liberated, and the higher the intensity of the source, the less time this process would take. Also, this implies that the kinetic energy of the photoelectrons should be a function of the light intensity (which is not what we see in experiement, in fact kinetic energy of photoelectrons are function of incoming light's frequency).
ok, given these observations, let's go back and talk about how does a classical wave transfer energy. Clearly if you stand under the sun for too long, you will "heat up"
. Why? the energy traveling from the sun in the form of EM radiation continuously strikes your body (the parameter here is in energy flow per unit area which can be described by the Poynting's vector) and you gradually absorbed more and more of it. If it is on a cloudy day (ie. less intensity), it will take longer for you to "heat up" to the same level. If you are an electron residing inside some metal, you will expect that the longer you are being shone on, the more energy you will absorb and eventually gain enough energy to get out of the metal.
A classic example of such "build up of energy" can be illustrated by a simple experiment like producing a sound wave of the correct frequency near a glass bowl/cup until it shatters. The correct frequency of course is the
resonant frequency (fundamental or its harmonics) of the bowl/cup. The loudness of the sound is what we called "intensity" of the wave (or amplitude). So, if you turn the sound on and eventually the glass bowl/cup will break (well... we did an experiment on this to show the students this idea, but we did choose a type of glass that wouldn't break as easy... so that we won't have to keep buying new glass for every class
, but one could clearly see the oscillation of the bowl slowly becoming more violent).
Anyway, the idea here is clear: energy from the sound wave is transferred to the bowl very effectively (because it is at resonant frequency), and the vibration builds up and ultimately breaking the bowl. Should the amplitude of the sound wave is bigger, this will happen much faster. Mind you, you may turn the amplitude down to very tiny value, but eventually it will gain enough energy to break it. This is an example of energy transfer via a classical wave.
The kind of calculations you may be thinking of here is firstly work out how much energy per unit area is being transferred per unit time, and then see how many electrons are there per unit area... plus few other stuffs like how does the electrons absorb and redistribute the energy within the lattice etc. All done assuming energy is transferred as a continuous wave medium. Not going to be an easy calculation, but since all we are after is the question: would any of the electron receive enough energy to be liberated? Therefore, from the glass bowl analogy we would expect the wave-model to predict that as long as we wait long enough, it will absorb enough energy.
Now, the fact that a wave-model seems to predict a different story for the photoelectric effect, it is an indication that another model is needed,.. it turns out that the particle model works fine. Note that the
resonant frequency is not like the threshold frequency we were talking about. The former comes in discrete values, while the latter is like a cutoff value above which all frequencies will work.
what would the particle model predict:
now, here each photon is like a billard ball carry energy defined by its frequency. When such billard ball of enough energy strikes an electron, energy is transferred to from the photon to the e- and so the e- is liberated. The idea here is that if the photon does not carry enough energy, the e- will not gain enough to be freed. And it doesn't matter how many of such photons are around, you still can't do it. But why? couldn't you have two or more of such low energy photons hitting the same electron in a short space of time (so that the e- hasn't lost the energy from the previous photon strikes through vibration in the lattice) and hence eventually building up enough energy just like in the wave-model? The difference here lies on the fact that it is now a discrete source rather than a continuous source like in the wave-model. So, the calculation can now become a bit more technical where you will have to take into account of the
probability of hitting the same e- within a certain time frame. Needless to say some ppl have tried doing this calculation before (with various assumptions, otherwise it is probably intractable), and as far as I was told by the lecturer I was working with, the probability that this mechanism can successfully build up enough energy for a particular e- is extremely low. In fact on average, you would have to wait longer than the age of the universe to see it.
The moral is that in the photoelectric effect, light behaves more like a particle than a wave. And the existence of a threshold frequency (and other observations) is indicative of that.
