3 ms·

If I emit a bass signal at a low amplitude, but then emit it at a higher amplitude, I can see the effect on a glass of water on the table. What’s happening here

• tim-kt 2y ago
  Sorry, my last sentence wasn't formulated well. Yes, a wave with higher amplitude (or one could say "intensity") has a higher energy. The photoelectric effect happens when you shine light with "enough" energy on some material such that the atoms of the material are ionized, i. e. electrons are freed. You need a minimal energy for this and if you use dim light with a low frequency, you will not see the effect. Now, if you increase the frequency of the light, you can measure electrons. If, instead, you make the light brighter, that is, increase the amplitude of the wave (if it were a wave), you don't see electrons. So at least in this experiment, light does not function as a wave.
  • deleted 2y ago
    [deleted]
  • HarHarVeryFunny 2y ago
    But once you've increased the light frequency (i.e. photon energy) above the required threshold, THEN making the light brighter (more photons) will increase the number of electrons emitted.
• tsimionescu 2y ago
  The point here is that the total displacement of water caused by a sound wave depends on both the amplitude of the wave, and its frequency, with no limit: if the wave has high enough amplitude, it will displace water even if the wave is very low frequency. However, this is not true for EM interactions. If you shine infrared light on a solar panel, you'll see 0 current from it, even with an extremely powerful source of light (at some point the material might heat up enough it starts showing some thermo-electric effect, but that's a different thing). However, if you take even a very low intensity ultraviolet source, you'll see a measurable current right away. This is the unexpected behavior that quantized interactions have, which can't be reproduced with non-qunatized waves like sound waves.