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What is it about the molecular structure of dark coloured objects that means they absorb most of the visible light that falls on them, but radiate infrared?

Naively I would think that if a molecule can absorb visible light, it could emit it just as easily. The background is this post in earth science Does visible light warm the earth?, which is clearly more a physics question.

What is going on at the microscopic level?

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    $\begingroup$ Depends on what temperature they are. See black body radiation. $\endgroup$
    – Jon Custer
    Aug 16 at 15:03
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    $\begingroup$ If you shoot a football into a net, the ball stops. But any objects that were in the net might fly out of the net. That is basically what happens. If you shoot the ball at a concrete wall, the ball returns and the objects on the wall remain. It is all about the energy levels in the shot versus the rigidity of the net. $\endgroup$ Aug 17 at 8:15
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    $\begingroup$ Related: physics.stackexchange.com/q/645671/247642 $\endgroup$ Aug 17 at 8:27
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    $\begingroup$ Broadly because that's what it means for an object to be "dark" $\endgroup$ Aug 17 at 22:03
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    $\begingroup$ This is true for any object at room temperature, no matter the color. A blue object at room temperature does not "radiate" blue light. It reflects blue light from an external source. But its radiation is still in the infrared or below (in frequency). $\endgroup$
    – nasu
    Aug 18 at 12:02

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At the microscopic or molecular level you have electronic transitions and molecular transitions. Electronic transitions for example the electron is excited to a higher energy level. In a molecular transition the molecule in a spring model (bonds like springs between atoms - phonons) the molecule gets excited to a higher mode of vibration. For electronic transitions they are often in the visible spectrum, molecular transitions they are in the infrared, or far infrared. You can also look at semiconductors where instead of discrete energy levels you have different energy bands. Usually we talk about a conduction band or a valence band, but there are higher empty bands that electrons could get excited to, or when an electron is excited to conduction band it may be excited to a state that is not at the minimum energy of the conduction band. For metals the absorption might be from collective oscillations of electrons called plasmons.

Anyway the point is that if you look at it from an energy level perspective and you excite to a higher level. The absorption excites the particle (electron, phonon, etc.) from a lower energy to a higher energy and you could have re-emission from that energy state, but often there are other processes where that particle loses energy through other processes before it can re-emit.

That loss of energy typically ends up in the form of heat, but if you wanted to look at it from a particle point of view that "heat" might be a phonon vibration that goes through the lattice.

If it was a solar cell - sometimes you might thing of this as a loss of efficiency, a high energy photon comes at 2.5 eV, but the band gaps of the silicon is about 1.1 eV, the photon gets absorbed losing 2.5 eV, when the electron rattles between allowed states down to the bottom of the conduction band it only has about 1.1 eV - so about 1.4 eV was lost in the form of heat. Silicon is also not very good at emitting light so it doesn't lose that energy in the form of an emitted photon.

If it was something like a dye molecule in a liquid and you illuminated it with visible or UV light, it might very efficiently emit photons between some energy levels, but usually that emission will be substantially longer wavelength than the exciting photon, and the difference in energy is going somewhere by vibrating the molecule or being transferred by molecule to molecule collision to the liquid solvent. Or if the dye molecule doesn't emit in the form of a photon it is heated up.

So if visible light from the sun - black body spectrum about 5700k some gets reflected, some gets absorbed, some might be remitted as photons at lower energy, but in general what absorbed is heating up the material and that material is at a lower temperature say 300K and radiates with a black body spectrum around 10 um. If the sun heats it up to 310 K, then the peak of the spectrum would emit at a slightly shorter wavelength.

Since materials can be complex it is often convenient to measure a emissivity or absorptivity when discussing how well the material emits absorbs or emits the thermal radiation.

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When a photon interacts with atoms or molecules of a gas, an electron goes to a higher energy orbital, and later on releases that energy. As $\Delta E = h\nu$, the emitted frequency matches the absorbed one.

The electrons of a solid are arranged in band structures instead of the orbitals of an atom. Each band has a lot of states with small energy and momentum difference between them.

An interaction can displace an electron of a band to a superior one, with more energy, if that band is empty, or at least has many available states. Once there, it is possible for the electron to reach a lower energy state, on that same band. And that $\Delta E$ is not necessarily the same of the interaction. The emitted frequency is different as a consequence.

Sometimes, there is an energy gap between bands, that is greater than the energy of photons of visible light. They don't interact in that case, making the material transparent, as is the case of quartz.

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  • $\begingroup$ I feel this best answers the precise question in a crowded field of very helpful posts, so have marked it as the accepted answer $\endgroup$
    – Peter A
    Aug 17 at 7:56
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I'm not a physicist, but here is what I recall (greatly simplified, not going into the quantum-physical causes – see other answers for that):

  • All frequencies of radiation, including visible light, are absorbed by a black body, and contribute to heating the body.

  • What frequencies are emitted by a black body (as well as the intensity) depends on its own temperature. (For higher temperatures, the emission curve is higher, and also has its peak more towards the higher frequencies/lower wavelengths.)

    Wikipedia has this diagram (ignore the black line):

  • For a typical objects around us, the temperature is just not high enough to create (measurable amounts of) visible light, but is high enough to create infrared light.

  • If you heat a black body high enough, it starts emitting light in the visible frequencies (first red, then shifting towards blue as we get hotter). (You might remember seeing hot coal in your barbecue grill, for example.)

An example of an everyday black body: Our Sun's photosphere has an average temperature of 5772 K, so it is emitting visible light (as well as ultraviolet and infrared).

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    $\begingroup$ A practical reversed example of this is blacksmiths determining the temperature of the material they are working with by how "red" or how "white" (=blue) it is. $\endgroup$ Aug 18 at 6:52
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Your naive expectation is actually exactly correct. See: Kirchhoff's law of thermal radiation and Einstein coefficients. Every photon a body emits it is equally capable of absorbing, and vice-versa. It's just a question of what it is likely to do, given the circumstances.

So, what's happening is you're observing an environment bathed in radiation from a hot body ($T_\odot \sim 5,500\,\mathrm{K}$) that produces copious visible light, surrounded by a cold blackbody ($T_{\mathrm{CMB}} \sim 3\,\mathrm{K}$), and at a middling temperature ($T_\oplus\sim 300\,\mathrm{K}$). If the objects around you were at $5,500\,\mathrm{K}$ they'd produce visible light in similar quantities to the sun. The details at any given wavelength would be different, because the albedo varies as a function of wavelength in a way that depends on the composition of the body (white things would emit less than black things), and we're not composed of ionized hydrogen, but broadly this is what's going on.

tl; dr: dark objects emit and absorb light at all wavelengths, they just emit very little visible light because their temperature is low, because visible light is on the Wien's law tail of its emission spectrum, which is an exponential cutoff.

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There is no such restriction. Infrared is just the peak of the radiation spectrum of objects at the temperatures naturally occurring in environments not rapidly fatal to humans. Shine a powerful industrial laser at a sheet of steel in a factory, and the steel will glow bright yellow-white (and melt) at the point of irradiance. An inexpensive mid-sized lens can concentrate visible sunlight sufficiently to heat objects until they glow red-hot (although you'd need to quickly shade them after heating them to see it over the bright reflected sunlight).$^1$ Shine a dim flashlight at a space rock in the frigid darkness of intergalactic space, and with sufficiently sensitive instruments you would find it radiating back a faint spectrum with a peak in long-wave radio and almost nothing in the infrared.

1: you can do this experiment in your back yard on a sunny day for about $20 for a mid-sized fresnel lens - but be very careful, as even the reflection of sufficiently concentrated sunlight could cause retina injury.

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My best answer is Entropy. First of all the amount of energy a body can reflect back is bound from above by the energy it receives, and we must recall that the energy of a photon grows with its frequency. Now if entropy is to increase from this process then the radiation that must come out of the object must have more degrees of freedom than the incident one since entropy is just the log of number of microstates. What this means is that for the entropy to increase the emitted light must have more photons, since every photon has just 2 degrees of freedom. Then for entropy to increase the number of emitted photons must be larger than the number of incident ones, since there are more photons in the emitted light but the same energy to share this means that each must be less energetic, more redshifted.

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  • $\begingroup$ Thank you, v helpful. I am leaving the question open to see if someone can add something on what happens at the microscopic level $\endgroup$
    – Peter A
    Aug 16 at 15:32

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