4
$\begingroup$

Why do hot objects prefer to emit photons over electrons ? Is there electron-positron annihilation ? If so , why ? Im confused by this.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
5
$\begingroup$

The energy levels in any object are quantized. The ground state of an electron is the lowest energy. From there electrons can have many higher energy levels, the highest being that which allows them to completely escape the nuclear attraction, i.e., get emitted.

So hot objects can emit electrons, but the probability that any single electron will have that much energy just due to heating the object is lower than electrons going up a few levels (due to gaining thermal energy) and then jumping back and emitting photons. Very hot objects can become plasma (electrons are emitted and you are left with ions). There are of course other ways to achieve a plasma (e.g., dielectric breakdown discharge) which are not equilibrium thermal excitation. But that is beyond scope of this question.

Improve this answer
$\endgroup$
5
  • $\begingroup$ But if the materials have electrons in the outher shell , they should be emitted ? But that is not light ? $\endgroup$
    – mick
    Commented Jan 21, 2013 at 17:28
  • 1
    $\begingroup$ @mick light is electromagnetic radiation, photons. electrons are particles, stable until they meet a positron. Positrons to be created in an electron positron pair need large energy photons,; these energies (MeV) do not exist in "hot" objects. $\endgroup$
    – anna v
    Commented Jan 21, 2013 at 18:11
  • $\begingroup$ @mick Strictly speaking there is no outer shell. There are a countably infinite shells but after a point the electrons are so far from the parent nucleus so essentially they are considered emitted. So the question is for a certain body temperature how far the electrons go. Also it is important to remember a hot body is not one molecule. If you have a solid with many molecules what you are seeing is lot many electrons with lot many energy levels (bands). $\endgroup$
    – Sankaran
    Commented Jan 21, 2013 at 18:26
  • $\begingroup$ contd... It is more likely that many will excited with moderate energies and fall back emitting and fewer will gain large energies to escape the material. But yes,thermo-electric emission is non-zero, it just has lower probability. $\endgroup$
    – Sankaran
    Commented Jan 21, 2013 at 18:26
  • 1
    $\begingroup$ good point, it requires more energy to make an electron completely leave the nucleus' electric potential compared to exciting an electron to 'just' a higher energy level from where it can 'relax' again to the original level by emitting a photon. $\endgroup$
    – Andre Holzner
    Commented Jan 21, 2013 at 20:28
5
$\begingroup$

Electrons are emitted, but being charged, they form a cloud close to the surface and do not travel too far due to attraction to the positively charged remainder. If you place a metallic plate near the surface, you can collect these electrons and even make a circuit. This is known as thermo-electric sources of electricity.

Photons are neutral and are not attracted to the hot body, so they go away.

Edit: There are no positrons there, but positively charged atomic ions. In a hot object there are processes of atom excitation/ionization and recombination. The latter process produces photons.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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