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The photoelectric effect [0] can be explained if light behaves as discrete particles, but not when it's a wave since a higher amplitude does not imply a higher
by tim-kt 2y ago
The photoelectric effect [0] can be explained if light behaves as discrete particles, but not when it's a wave since a higher amplitude does not imply a higher energy transfer.
[0] https://en.m.wikipedia.org/wiki/Photoelectric_effect https://en.m.wikipedia.org/wiki/Photoelectric_effect
- rusk 2y agoIf 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 if amplitude does not carry power? My understanding is that theoretically energy transfer is a function of wavelength.
- tim-kt 2y agoSorry, 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 agoBut 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 agoThe 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.
- HarHarVeryFunny 2y agoOK - a bit like the fairground game of trying to knock coconuts off a stand by throwing a wooden ball at them. It doesn't matter how many balls you are throwing per minute (total energy being delivered) if the energy of each ball doesn't cross the threshold to knock the coconut off. OTOH, the energy of a photon is such an abstract concept (not like the kinetic energy of a ball) that I'm not sure it really helps explain it.
- tim-kt 2y agoWell, there is a saying about spin of an electron. Imagine that you have a ball and it's spinning. Except, it's not a ball. And it's not really spinning.
- wasabi991011 2y agoYou can explain the photoelectric effect with classical light (i.e. as EM waves) as long as you properly quantize the atomic energy levels. This is often called s semi-classical model. However, photo-detections with sub-poissonian statistics cannot be explained under this semi-classical model, but it can be explained with properly quantized EM field (i.e. with photons). For reference, see Mandel and Wolf's Quantum Optics textbook.
- GoblinSlayer 2y agoParticles aren't necessary for it, any quantization is sufficient.
- tim-kt 2y agoSure, I'm using the word "particle" loosely.