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The other answers posted here are not quite right. Light energy is indeed inherently quantized, and a photon is one quantum of light. In other words, you can't
by URSpider94 10y ago
The other answers posted here are not quite right. Light energy is indeed inherently quantized, and a photon is one quantum of light. In other words, you can't have a half-photon of light, only even multiples.
Einstein wrote the equation "E = h*(nu)", where h is the so-called "Planck's constant", and nu is the frequency of light. Translated, this means that each photon carries an amount of energy proportional to its frequency, higher-frequency photons (IR -> red -> blue -> UV -> X-ray) carry more energy.
tl;dr: Quantization is an inherent property of light.
- DigitalJack 10y agoDo you mean "whole" multiples?
- URSpider94 10y agoYes, integer multiples would be better. Don't know why I said even.
- glitchc 10y agoE=h*nu was derived by Max Planck, not Einstein.
- URSpider94 10y agoDoh. Typing too fast. Thanks for the correction.
- amelius 10y ago> E = h*(nu) Yes, but the question is whether that is due to the material not being able to produce non-quantized photons. Stated differently, suppose we had a different way of generating photons, then could we theoretically create them in a non-quantized way?
- lisivka 10y agoYep. They called "radiowaves".
- amelius 10y agoSounds logical, but then again, how fundamental is the "Q" in "QM"?
- lisivka 10y agoExtremely. Broken quanta will release its energy. Check solitons (example: https://www.youtube.com/watch?v=JD32kkoFU3Y https://www.youtube.com/watch?v=JD32kkoFU3Y ). Soliton wave can cross ocean, but if you broke it pattern, you will have regular waves, not a half of solition.
- URSpider94 10y agoNo, radio waves are also made up of photons that obey the same equation.
- URSpider94 10y agoThe best example I can come up with is Compton scattering. Here, you are colliding photons with an unbound particle, therefore one that does not have quantum energy states. The quantum nature of light is shown by the fact that the degree of scattering depends upon the energy (and therefore wavelength) of the incident photon, and not on the intensity. If light were just a wave, then the effect would go away if the intensity were low enough.
- jkn 10y agoI don't think this is right. The energy of a photon is h multiplied by the frequency, but the frequency itself can be anything. Even if the process creating the photon is inherently quantized (e.g. bound electron with quantized energy levels), we will measure slightly different frequencies (and so different energies), due for example to the Doppler effect[1]. Still with the Doppler effect, you can give a photon an arbitrary frequency (energy) by changing the velocity of the reference frame where you make the measurement (that is, until someone shows that the velocity of the reference frame is quantized... not something that is presently known). [1] https://en.wikipedia.org/wiki/Doppler_broadening https://en.wikipedia.org/wiki/Doppler_broadening
- URSpider94 10y agoYes, that's correct. This does not mean that the energy levels of light (more generally, EM radiation) are quantized -- photons can have any energy level. What it means is that, for light of a given frequency, you can deposit energy only in units of h(nu). The quantization of light shows up in how it interacts with particles -- even unbound particles like free electrons, which also do not have quantized energy levels. Specifically, if light were NOT quantized, you could get the same effect with more intense light that you get with more energetic light. Instead, experiments show again and again that longer-wavelength light at high intensity gives a totally different effect from short-wavelength light at low intensity. Postulating that light consists of particles (photons) with E = h(nu) explains this difference.