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Physics
Classical Physics
Electromagnetism
How a solid body emits a lower frequency photon than absorbed
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[QUOTE="FrankJ777, post: 6060862, member: 125080"] I’m trying to understand how a solid body changes the wavelength of radiation it re-radiates from that which it originally absorbed. I’m thinking in context to the way that the Earth absorbs higher frequency radiation from the sun, but when it re-emits the energy it’s at a much lower frequency. What I *think I understand is that; approximating Earth and the sun as black bodies; the sun being at a much higher temperature radiates energy where the peak of the distribution is at a high frequency. The Earth absorbs the higher frequency energy and becomes “warmer”. When the Earth re-radiates the energy, because it’s at a much lower temperature, the distribution of radiation has a peak at a much lower frequency (IR). But also I thought that heat radiation is transferred at discrete energies. Where IR radiation is absorbed in a molecule by causing rotation and vibration, and re-radiated at the same frequency. While higher frequency radiation is absorbed by promoting an electron to a high state, and when it falls back to ground state it emits a photon of the same energy/frequency. In solids I *think that electron energies exist in continuous bands, but I thought that the same would apply as in a gas, where a photon with energy [B][I]hv[/I][/B] of visible light promotes a single electron to a higher state, exactly [B][I]hv[/I][/B] greater, and then it emits another photon also with energy [B][I]hv[/I][/B]. So how is radiation ever “converted” from high frequency light to IR? I suppose that the IR is radiated due to rotation and vibration of molecules? ... So is the question I should be asking is “how is visible light converted to kinetic energy? [/QUOTE]
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How a solid body emits a lower frequency photon than absorbed
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