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.

  • $\begingroup$ Which texts did you read? $\endgroup$
    – my2cts
    May 26, 2021 at 15:19
  • $\begingroup$ I read it on Khanacademy $\endgroup$ May 26, 2021 at 15:22
  • $\begingroup$ Try en.wikipedia.org/wiki/Photoelectric_effect. $\endgroup$
    – my2cts
    May 26, 2021 at 15:38
  • $\begingroup$ @my2cts Current does not increase with increasing kinetic energy per electron unless that energy goes toward dislodging additional electrons. Current is electrons per second, it doesnt matter how much energy you give them the generation rate remains unchanged. physics.stackexchange.com/q/222359/29170 $\endgroup$
    – Matt
    May 26, 2021 at 16:36
  • $\begingroup$ @Matt You are right . In this case the density varies inversely with velocity. $\endgroup$
    – my2cts
    May 26, 2021 at 19:30

2 Answers 2


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.

  • $\begingroup$ The current is determined by the number of electrons and their speed. $\endgroup$
    – garyp
    May 26, 2021 at 17:10
  • 1
    $\begingroup$ @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. $\endgroup$ May 26, 2021 at 17:16
  • $\begingroup$ 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. $\endgroup$ Jun 2, 2021 at 5:21
  • $\begingroup$ @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. $\endgroup$ Jun 2, 2021 at 10:53

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

  • $\begingroup$ 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? $\endgroup$
    – garyp
    May 26, 2021 at 17:16
  • $\begingroup$ It gives higher KE instead, I think $\endgroup$ Jun 2, 2021 at 5:13
  • $\begingroup$ 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). $\endgroup$ Jun 2, 2021 at 10:59
  • $\begingroup$ @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. $\endgroup$
    – stuffu
    Jun 2, 2021 at 16:03
  • $\begingroup$ @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. $\endgroup$
    – stuffu
    Jun 2, 2021 at 16:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.