5
$\begingroup$

This is my understanding of the black body radiation: a black body is heated. The energy from the heat causes the electrons in the body to oscillate, so they will emit electromagnetic waves. If we plot the intensity of radiation against the emitted frequencies, we'll see that beyond a certain frequency, the intensity is decreasing for higher frequencies. Here are my questions assuming the above description is correct:

  1. Is it true that there are equal number of electrons emitting each frequency or there are less electrons emitting higher frequencies?

  2. Is intensity of radiation related to a wave's amplitude? or is it related to number of electrons emitting a certain frequency?

  3. If there are equal number of electrons emitting each frequency, is it then the amplitude of the waves which demands a higher frequency wave carrying more quanta of energy?

$\endgroup$
1
$\begingroup$
  1. Higher energies are less probable, so there are less electrons emitting high frequency photons.

  2. The intensity is proportional to amplitude squared, in the wave description, and proportional to number of photons in the particle description

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ another question: the intensity obeys Boltzman's distribution. Is there a Brownian motion in this process? if yes between which particles? I think it cannot be between atoms or electrons since they do not collide randomly. Or is it just a random distribution of energy between atoms which should obey the probability law behind Boltzman's distribution? $\endgroup$ – kave33 Jul 23 '16 at 19:16
  • $\begingroup$ Brownian motion is motion of larger (still microscopic) particles in a fluid, when random atoms collide with that particle. The second answer is correct $\endgroup$ – Andrei Jul 23 '16 at 20:17

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.