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I read in some texts that classical physics predicted the following in the photoelectric effect,

  • KE of electrons ejected is directly proportional to intensity of light
  • Increasing the frequency would increase measured current. this is the text I am referring to

My question's straight, why was it that they predicted such an effect of varying frequency for the Photoelectric effect? And also give an example of the effect of frequency on an experiment such as jerking a rope with a high frequency and transmitting the energy onto a beach ball, which is based on the wave-physics of the classical physics.

Lumbini A Tambat
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Presumably you know that experiments demonstrated the falsehood of the Classical model.

The model was based on the simple idea that the more energy you hit the electrode with, the more energy it would give each electron. Light was treated as a wave of such energy.

In that model, it follows that greater wave amplitude (light intensity) would impart more energy to everything it hit.

It is less obvious that a shorter wavelength carries greater energy than a longer wavelength of the same amplitude. It has to do with the rate of change of the wave form (the slope of the curve when you draw it), and is true in both the classical and quantum models.

However I am unclear why shorter waves should be expected to increase the current, as that is the number of electrons not their energy. Perhaps it is because there are more peaks per second, which would supposedly therefore knock more electrons out.

Of course, we all know that experimental observations gave the lie to all that. Einstein explained it as energy thresholds, by treating the light as discrete particles or packets, and in doing so co-founded quantum physics and earned himself a Nobel prize.

Guy Inchbald
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  • The current is determined by the number of electrons and their speed. – garyp May 26 '21 at 17:10
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    @garyp Current is charge per second. That does not change just because the charges change their velocity. Think number of cars per minute passing a radar speed check vs. number entering the roundabout at the next junction. You appear to be getting the number emitted confused with the electron density. – Guy Inchbald May 26 '21 at 17:16
  • It would be really grateful of you could briefly explain why for the same amplitude, lower frequency gives higher energy in the wave nature model of classical physics. – Lumbini A Tambat Jun 02 '21 at 05:21
  • @LumbiniATambat It's the other way round. As the answer says, "a shorter wavelength carries greater energy." A shorter wavelength corresponds to a higher frequency, not a lower one as you suggest. It is as true of ocean waves or sound waves as of light quanta. If you need to know more than the answer gives, best to treat it as a different question from the one here. If SE has not already been asked and has answered, you could ask it as a new SE question. – Guy Inchbald Jun 02 '21 at 10:53
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Beach ball A and beach ball B are in a depression on sand. Ropes are connected to the balls. A low frequency is sent through the rope to ball A. A high frequency is sent through the rope to ball B.

Ball A sways back and forth, then leaves the depression. The kinetic energy of ball A is random value between zero and the energy of the last wave on the rope.

Ball B sways back and forth, then leaves the depression. The kinetic energy of ball B is random value between zero and the energy of the last wave on the rope.

The energy of the aforementioned last wave is proportional to the amplitude and the length of the wave. Or was it amplitude squared? Anyway, we have decided that the amplitudes are the same, right? So the maximum kinetic energy of ball A is 3 times larger if the last wave absorbed by ball A is three times longer than the last wave absorbed by ball B.

Here "last wave" means the wave-crest that causes the ball to leave the depression.

Energy available for dislodging balls = Total available energy - kinetic energy of free balls

stuffu
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  • It seems to me that according to this model, the wave with the lower frequency gives the ball more energy. Applying the model to the photoelectric effect, the model is saying that lower frequency gives higher current ... or am I missing something? – garyp May 26 '21 at 17:16
  • It gives higher KE instead, I think – Lumbini A Tambat Jun 02 '21 at 05:13
  • It is wholly wrong to suggest that "The energy of the aforementioned last wave is proportional to the amplitude and the length of the wave." The energy in each wave is proportional to the slope of the wave, which is inversely proportional to wavelength. Given a ball which is relatively small and light, the short wave (a sharp jerk) will eject it sooner than a long wave (steady tug). – Guy Inchbald Jun 02 '21 at 10:59
  • @GuyInchbald Nah, slope is not important. Like for example small ripples in a pond have very little energy. Height is important. Water high up has a lot of potential energy. – stuffu Jun 02 '21 at 16:03
  • @garyp Current is number of charged particles per second. Current is charging speed. Electrode from which electrons fly off is being charged. At rate n electrons per second. Which is same as x amperes. – stuffu Jun 02 '21 at 16:11
  • @stuffu slope is proportional to frequency and to amplitude. Low-amplitude waves have shallow slopes unless they are very high frequency. The relationship between energy, frequency and amplitude for water surface waves is very different from that for electromagnetic or spatial acoustic waves. Don't be fooled by spurious analogies. – Guy Inchbald Jun 02 '21 at 17:19
  • @GuyInchbald Hmmm.... when the guy starts making those short waves he should take a step towards the beach ball, otherwise he would increase the tension of the rope. If the guy does that, then I believe the energy of a wave crest is proportional to amplitude squared. Well I guess you disagree. Maybe I can find in Wikipedia something that supports my position. Oh and the rope should be bendy, I mean not resist bending. – stuffu Jun 02 '21 at 17:56
  • @GuyInchbald the rope-wave we maybe could forget, but this looks very relevant: https://physics.stackexchange.com/questions/640753/em-wave-power-dependency-on-frequency – stuffu Jun 03 '21 at 02:11
  • @stuffu OK, it depends how you define "amplitude". If you just mean field strengths, you are correct about frequency independence. But if you mean photon density, then a short-wave photon has higher energy than a long-wave photon, and there are fewer of them for the same power. Perhaps I should have made that clearer. – Guy Inchbald Jun 03 '21 at 08:42