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I'm learning about the photoelectric effect and blackbody radiation separately at school. However, I'm having trouble reconciling the two.

From what I know about blackbodies; they are perfect absorbers and emitters of electromagnetic radiation. Due to the random movement of charged particles within the blackbody, electromagnetic waves are emitted.

However, I also know of the photoelectric effect which occurs when electromagnetic radiation is shone onto a metal surface. Since metals have free valence electrons, the electrons absorb photons of energy and are liberated.

Are these two effects related? From my understanding, the emission of blackbody radiation is not the same as electrons being emitted due to the photoelectric effect. In that case, can a blackbody exhibit the photoelectric effect?

Any help will be greatly appreciated, thank you.

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3 Answers 3

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The reason quantum mechanics became a necessary evolution of the theory of physics is based on three very important experimental effects, that could not be mathematically modeled with classical electrodynamcis, i.e. Maxwell's equations.

  1. The atomic spectra . Instead of a continuous frequency specrum, there were absorption and emmission lines.

  2. Black body radiation in classical electrodynamics could not be modeled, there was the ultraviolet catastrophy

  3. The photoelectric effect could not be explained with classical EM, because it was expected that the stronger the light was, the higher the current in the experiment. Instead the current was found to depend on the frequency of the light, with a cutoff.

The hypothesis that classical electromagnetic radiation was composed of quanta called photons, with energy $=hν$ where h is Planck's constant and $ν$ the frequency of the classical electromagnetic wave explained all three puzzles , and finally led to the Bohr model and Schrodinger's equation, the beginning of quantum mechanical theory.

The effects are related to the successful hypothesis of classical electromagnetic waves being a superposition of photons.

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  • $\begingroup$ Thank you for this. I understand how they fit into the bigger picture of how the light model was changed, but I'm still having trouble understanding the actual physics itself. For instance, if you had a blackbody with electromagnetic radiation shone upon it, the nature of it being a blackbody means that it will absorb all radiation. So some electrons would jump energy levels and leave, hence would it experience the photoelectric effect? In that case, blackbodies can experience the photoelectric effect? Thank you $\endgroup$
    – latin333
    Jun 22, 2021 at 5:06
  • $\begingroup$ Black body radiation is the photons leaving the volume of the body, though created within the body from scatterings the energy provided by the kinetic energy modeled by the temperature. In a cavity the formula is true. In a body it has been found to be approximately true, see the sun spectrum. arxiv.org/abs/2103.12724 There the energy of the photons is enough to have internal interactions and change the spectrum (yellow curve) from the formula. Now in a conductor, the photo electric effect appears from shining photon beams on the surface. $\endgroup$
    – anna v
    Jun 22, 2021 at 6:50
  • $\begingroup$ At normal temperatures the black body photons do not have enough energy to change energy levels in the volume, and even if, from qm probability, one could do it, it will be very very rare, even in the volume. To reach the surface of a conductor is even more improbable. $\endgroup$
    – anna v
    Jun 22, 2021 at 6:55
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In the photoelectric effect, electrons are 'kicked out' from the surface of a metal. The kinetic energy of those electrons depends on the frequency of the light shone on the metal.

This supports the idea that the energy of arriving photons depends on frequency $E=hf$ and that the energy is arriving as discrete 'quanta', small packets - the electrons are ejected as soon as the light is shone on the metal and not after a delay as predicted by classical theory.

Blackbody radiation is about the distribution of frequencies of photons that are emitted, so it's the photons that are emitted in this case - whereas in the photoelectric effect it's the electrons that are emitted.

The two are connected as they are both topics that needed quantum theory to explain. Classical theory didn't explain the photoelectric effect. Also using classical theory there was a problem with black body radiation, the energy predicted to be emitted at higher frequencies was too high compared to experiment and could approach infinity - the ultraviolet catastrophe. Max Planck with the quantum hypothesis, was able to explain the blackbody radiation distribution.

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  • $\begingroup$ Thank you, the distinction between photon and electron emission is really helpful! In that case, my understanding is that electrons in the object are randomly moving thus releasing electromagnetic radiation/photons. However if radiation is shone onto these electrons at the required frequencies, they absorb this energy and are completely ejected from the object's surface altogether (describing photoelectric effect). $\endgroup$
    – latin333
    Jun 22, 2021 at 5:10
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There really is no direct connection other than they are both phenomena that are used to demonstrate the quantised nature of light in terms of photons of discrete energy.

Blackbody radiation is not a mechanism in itself. The photoelectric effect could be a contributor to the necessary absorption processes required for something to be considered an approximation to a blackbody. If that were the case then the rate at which electrons were being liberated would be in equilibrium with the inverse process - known as photo-recombination, where a free electron is captured by an ion (or atom) and a photon is emitted.

In general, something emitting like a blackbody will be experiencing a variety of radiative processes. The photoelectric effect might be a contributor to this at short visible, UV and X-ray wavelengths, where the photons have the ability to liberate electrons from atoms and ions and photo-recombination could supply visible, UV and X-ray photons.

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  • $\begingroup$ I'm still a bit unsure as to how the two are related. Does that mean that emitting electrons is a form of radiation on the electromagnetic spectrum? $\endgroup$
    – latin333
    Jun 21, 2021 at 13:37
  • $\begingroup$ @latin33 photoelectric absorption is an absorption process. The inverse process (called photo-recombination) is one way that photons get emitted. $\endgroup$
    – ProfRob
    Jun 21, 2021 at 15:08
  • $\begingroup$ Ok, so if my understanding is correct: all objects are always releasing radiation (which is not the same as blackbody radiation). This is unrelated to the photoelectric effect, wherein electrons are liberated from the surface of metals by absorbing EM radiation. So therefore, blackbodies can experience the photoelectric effect. Thank you for your patience! $\endgroup$
    – latin333
    Jun 22, 2021 at 5:00

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