So a blackbody absorbs all incoming radiation and at thermal equilibrium emits radiation according to Planck's law. I have several questions.

How could blackbody absorb all incoming radiations? In order to absorb radiation, electrons would have to jump to higher energy levels or ionize and once an atom reaches that state it wouldn't have the same set of "jumps" left to continue absorbing as efficiently at that wave length. So its absorption rate would eventually change.

Also at the higher energy levels like gamma rays, it doesn't seem likely that a blackbody can constantly absorb gamma ray. Eventually it would be really ionized until the energy cost of more ionization is greater than the energy of the photon at which point the gamma ray must pass through?

At thermal equilibrium, the blackbody is absorption and emission rates are the same. But why does the emission spectrum follow that particular shape (I know it's empirically observed but are there some models that explain it). I thought emission spectrum is described by the distribution of electron movements. Why would that distribution always be the same at a certain temperature? Shouldn't it depend on the energy that is absorbed in some way?



Atoms with well-defined emission and absorption lines, e.g. in a single atom or ideal atomic gas, are _ far_ from being a black body. They’re just not the right model to understand continuous absorption and emission.

For that, you want to consider something with continuous energy spectra: free electrons, a metal, an ionized solution or gas, etc. There any energy state is available, so any energy can be absorbed or emitted.

Of course, most real situations have some of both properties: some continuous emission and absorption, but also lines and bands. Then it can get complicated to do detailed work.

  • $\begingroup$ There are no well-defined emission/absorption lines in crystals (well, if we don't count excitons and hydrogen-like impurities). The bulk of absorption and emission in crystals is organized in bands instead of lines. $\endgroup$ – Ruslan Dec 4 '19 at 9:10
  • $\begingroup$ @ruslan good point. Thanks. Fixed. $\endgroup$ – Bob Jacobsen Dec 4 '19 at 13:59

The answer is simply that if the material does not manage to absorb all radiation at all frequencies then it cannot be a perfect black body.

Why doesn't the absorption rate change as more photons are absorbed? Because it is in equilibrium. This is what equilibrium means - there are as many emission events as absorption events. The population of excited states and the distributions of particle velocities are determined by Boltzmann statistics and Maxwellian distributions at a particular temperature. If there is "too much" absorption, then the system will be out of equilibrium, absorbing heat and... getting hotter.

e.g. your gamma ray example. Yes, you are ionising material, but in an ideal black body in equilibrium the rate of recombinations, resulting in gamma ray emission, would exactly balance them. If not, then it's out of equilibrium and not a blackbody. (Note that blackbodies at gamma ray energies are found at the centres of stars).

  • $\begingroup$ Thanks for the answer. But in this ideal blackbody, why does the fact that it is a blackbody and temperature necessary determine its spectrum? Couldn't there be different blackbodies (they all absorb all radiations) that have different spectrums? I.e. why is the blackbody spectrum uniquely defined by its property that it absorbs all radiations. Is there a mathematical derivation? $\endgroup$ – Shuheng Zheng Dec 4 '19 at 18:37
  • $\begingroup$ @ShuhengZheng The radiation field must be at the same temperature as everything else. The derivation of the Planck function is a standard textbook item. $\endgroup$ – Rob Jeffries Dec 4 '19 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.