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I have read that heat radiation happens in the form of infrared, which is an EM radiation with a longer wavelength than visible light. So the heat radiation that you can feel in an oven or under the the sun is actually the infrared portion of the total radiation. This is why fluorescent or LED lights are so bright but they don't heat up a lot - they mostly produce radiation in the visible spectrum with negligible infrared, whereas incandescent bulbs used to produce a lot of infrared as a byproduct (some would say the visible light was the byproduct in this case).

My question is why does electromagnetic-radiation in some wavelengths heat things up, whereas others, with both longer or shorter wavelength (RF, Microwave, UV, Gamma), don't have the same effect? Is it because of the size of the atoms/molecules, or inter-atomic distance, or the distance between nucleus and electrons? Some wavelengths are better suited to increase the vibration of the atoms than others?

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    $\begingroup$ The premise is questionable: All absorbed electromagnetic radiation heats the absorbing matter (minus some chemical or nuclear reactions) -- what else could it do? (Here is a discussion how absorbed synchrotron radiation heats cooled probes.) It's just that there is so much IR in your examples that it dominates. $\endgroup$ – Peter - Reinstate Monica Jul 31 '20 at 15:08
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    $\begingroup$ Re, "heat...that you can fee...under the the sun is actually the infrared portion." If you put a completely black-at-all-wavelengths object out in the bright sunshine, about half of the heat that it absorbs will be from the visible portion of the Sun's light. $\endgroup$ – Solomon Slow Jul 31 '20 at 17:39
  • $\begingroup$ Actually, the heat you feel under the sun is mainly the visible light! LED lights are simply still pretty dim compared to the Sun. $\endgroup$ – Jan Hudec Jul 31 '20 at 21:24
  • $\begingroup$ @SolomonSlow No, by far the largest part of the heat will be from the visible part. $\endgroup$ – Karl Aug 1 '20 at 21:48
  • $\begingroup$ The premise if this question is simply wrong: All radiation that gets absorbed causes heating, and the shorter the wavelenght, the more. $E=h\nu$ en.wikipedia.org/wiki/Planck%27s_law $\endgroup$ – Karl Aug 1 '20 at 21:52
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In a solid, "heat" consists of random vibrations of the atoms in that solid around their equilibrium positions. If the radiation striking that solid has a wavelength component that is close to one of those possible vibration modes, then the radiation will couple strongly with that vibratory mode and the solid will accept energy from the incident radiation and its temperature will rise.

If the incident radiation has too high a frequency (X-ray or gamma) the coupling is poor and the radiation just goes right through without interacting much. If the frequency is too low (radio frequencies lower than radar) the radiation bounces off and also doesn't interact much. This leaves certain specific frequency bands (like infrared and visible light wavelengths) where the interaction is strong.

Note that this picture is somewhat simplified in that there are frequency bands in the gigahertz range where the RF energy bounces off electrically conductive materials like metal (this gives us radar) but interacts strongly with dielectrics and materials containing water molecules (this gives us microwave ovens).

Note also as pointed out below by Frederic, molecules possess resonant modes that their constituent atoms do not and these can be excited by RF energy as well. Many of these molecular modes lie within the infrared range, giving rise to the field of IR spectroscopy.

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    $\begingroup$ 2.4 GHz is a radio frequency (NATO E-band), yet quite capable of heating something. Ask your microwave if you don't believe me. Could you add further details to your answer to clarify? $\endgroup$ – Mast Jul 30 '20 at 13:43
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    $\begingroup$ This answer is good but I think it should additionally address the misconception about visible light not being heat radiation. A fire does emit visible light as heat radiation, and objects that heat up under visible light do exist: they're called black. $\endgroup$ – FrederikVds Jul 30 '20 at 13:56
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    $\begingroup$ @FrederikVds A fire emits visible light, but most heat is emitted as infrared radiation. I'm not sure about black objects, but they might be absorbing infrared, too. $\endgroup$ – smcs Jul 30 '20 at 14:05
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    $\begingroup$ @smcs My point is just that visible light is perfectly capable of transmitting heat. The reason we associate infrared with heat is because high energy visible light sources are very rare (because they're rare in nature, hard to make artificially, and unsafe for your eyes). The question seems to suggest that the writer believes visible light can't absorbed as heat though, which is a misconception. $\endgroup$ – FrederikVds Jul 30 '20 at 15:20
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    $\begingroup$ @smcs The vast majority od the sun's energy flux is in the visible band, not in IR. There is very little overlap between the thermal radiation of the Sun and of the Earth. That strongly simplifies many aspects of modeling the radiation processes. $\endgroup$ – Vladimir F Jul 30 '20 at 17:19
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As mentioned by niels nielsen, the EM waves get absorbed when their wavelength matches the vibrational modes of the atoms in the solid. This causes the atoms to vibrate even harder and thus rising the temperature. From the vibration of atoms in solids, the extension can be made towards the vibration of polymers and organic molecules, which have additional vibrational and rotational modes. For example, in organic molecules, the entire molecule can vibrate (in addition to the single atoms in it). This vibration happens on a different length scale and therefore EM radiation with different (lower) frequencies can get absorbed with respect to regular solid materials. Furthermore, also rotational modes exist in these organic molecules which can also absorb EM radiation and give rise to heating.

As humans are made of these organic molecules, it are these molecular vibrational modes that absorb the IR radiation and that give us the feeling of temperature.

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ALL electromagnetic waves transfer energy. When they meet some body, they are either absorbed, scattered, or partially both.

The part that is absorbed always heats the absorbing body. Absorbed EM waves may, or may not do other things as well (like chemical changes or electric currents).

The reason why you don't feel much heat from a LED bulb is because the LED bulb doesn't radiate much. An oven is 1-5kW, the Sun is some 1 kW / square meter at noon. A typical LED bulb is 3-15W.

You can try some powerful LED (there are e.g. 50W directed LED headlights that can burn your skin pretty much).

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    $\begingroup$ I wonder if you can focus one of those regular LED bulbs down well enough with a decent lens to burn you, I work with a 1532 nm laser at 400 mW that will definitely do it if you hold your hand in the beam for a few seconds while trying to adjust another mirror $\endgroup$ – llama Jul 30 '20 at 18:51
  • $\begingroup$ @llama Hardly. These diodes are 0.3-0.5W each and the light output is probably ~0.1W or less. The rest is heat that goes to the heatsink. You can focus it w/ elliptic mirror and probably feel it w/ some more sensitive part of the skin. $\endgroup$ – fraxinus Jul 30 '20 at 20:57
  • $\begingroup$ Ah right I thought you were talking radiant power, not total power $\endgroup$ – llama Jul 30 '20 at 22:02
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    $\begingroup$ You can pop balloons and burn things with visible-light laser diodes, like red lasers found in DVD burner drives and blue lasers found in Blu-ray writer drivers. $\endgroup$ – Nayuki Jul 31 '20 at 5:51
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    $\begingroup$ I can burn fine with under 1W LED, no need for more. It's a matter of focusing. $\endgroup$ – Overmind Jul 31 '20 at 11:14
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It is very important to understand that heat energy is stored in the degrees of freedom of molecules.

Heat energy, at a microscopic level, is stored in the degrees of freedom of atoms and molecules. These degrees of freedom are translational, rotational and vibrational. They all store different amounts of energy, depending on the geometry of the atom. Translational degrees of freedom are the atom or molecule moving around in space, and there are always 3 for the 3 dimensions of space. The rotational and vibrational modes come from the geometry of the atom/molecule.

How is heat represented on a quantum level?

Now there are mainly three types:

  1. translational

Translational degrees of freedom arise from a gas molecule's ability to move freely in space.

  1. rotational

A molecule's rotational degrees of freedom represent the number of unique ways the molecule may rotate in space about its center of mass which a change in the molecule's orientation.

  1. vibrational

The number of vibrational degrees of freedom (or vibrational modes) of a molecule is determined by examining the number of unique ways the atoms within the molecule may move relative to one another, such as in bond stretches or bends.

https://en.wikibooks.org/wiki/Statistical_Thermodynamics_and_Rate_Theories/Degrees_of_freedom

Now you are asking, why do certain wavelength photons heat up certain materials' molecules only while others cannot?

Every molecule has its own quantum mechanical characteristics, which includes the translational, vibrational and rotational modes' characteristics, and what wavelength photons those can correspond to. This means that certain wavelength photons energy needs to match the energy gap between those modes.

If the energy of the photon matches (or sometimes is exceeding) the gap between two modes, then the photon might be absorbed with high probability.

Now it is not just that simple. Certain wavelength photons do have the ability to transfer their energies with higher probability to molecules that have a certain type of available degrees of freedom (mode).

Thus, certain molecules that have degrees of freedom available in the different translational, vibrational or rotational modes, can be excited by different wavelength photons.

Just a note, the other answers do not address this, but heating a material is contrary to popular belief not only mainly by absorption. A lot of photons' energy is transferred by inelastic scattering. In this case, the photon does not cease to exist, and only transfers part of its energy to the molecule.

https://en.wikipedia.org/wiki/Inelastic_scattering

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Things heat up when they absorb radiation. They do not heat up if they are transparent to that radiation or if they reflect that radiation. When they are transparent, the radiation passes through without losing much energy.

Different wavelengths have different absorption profiles in different materials (owing to the materials' atomic structure) as mentioned by @nielsnielsen and @Frederic. Glass looks pretty transparent in the visible spectrum, but absorbs infrared and ultraviolet radiation, so it's opaque for these wavelengths. I used to operate an infrared spectrometer. Instead of using glass to mount the samples, we had to use disks made of salt, since salt remains transparent in the infrared range. Water also appears to humans quite transparent, but strongly absorbs microwaves (the basis for microwave ovens).

What I haven't seen explicitly in the other answers is that the radiation is absorbed at the atomic and molecular level when the photon energy ($E = h \nu = h c/\lambda_\mathrm{vacuum}$) is equal to the energy required for quantum transition between different modes. These modes can include electron transitions in atoms, transitions of delocalized electrons in molecules, vibrational transitions of atomic nuclei, rotational transitions of atomic nuclei, and displacements of atoms in crystals.

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    $\begingroup$ "...if they are transparent," OR if they reflect it. Wearing all white clothing on a bright sunny day will keep you significantly cooler than wearing all black clothing. $\endgroup$ – Solomon Slow Jul 31 '20 at 17:37
  • $\begingroup$ @SolomonSlow I think you are thinking of a shiny silver reflective clothing, that stay cool by reflecting. White absorbs and re-emits it, that's how it stays cooler, you are correct. $\endgroup$ – Árpád Szendrei Aug 1 '20 at 15:41
  • $\begingroup$ @ÁrpádSzendrei, I am not an expert in the field, but I thought that the process you described--absorbs and reemits--was also known as diffuse reflection. Reflections off of a shiny silver surface would be specular reflections. $\endgroup$ – Solomon Slow Aug 1 '20 at 22:28
  • $\begingroup$ @SolomonSlow " reflection of light or other waves or particles from a surface such that a ray incident on the surface is scattered at many angles rather than at just one angle as in the case of specular reflection." Diffuse reflection means that the microsurface is not plain (or is so that the angles will be random). $\endgroup$ – Árpád Szendrei Aug 1 '20 at 22:40
  • $\begingroup$ @SolomonSlow I added a mention of reflection. Thanks! I forgot about that. $\endgroup$ – WaterMolecule Aug 3 '20 at 14:51

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