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When electrons transition from a higher energy state to a lower energy state (energy difference $\Delta E$), they produce massless photon with frequency $\nu$ where

$ \Delta E= h \nu$

(h is Planck constant). We know energy-mas relation $ E=mc^2$. Why not create some kind particle, in this case a particle that has mass m that we could calculate from the energy difference of the two states of the electron? Is there any kind critical energy difference $\Delta E_c$ such that lower than $\Delta E_c$ always is creating photon and higher than $\Delta E_c$ its value create particle with mass?

baponkar
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3 Answers3

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There are a few reasons why the particle produced needs to be a photon. Aside from conserving energy, we also need to conserve momentum, charge and spin, for example. So you would need to ask what other particle, instead of a photon, could be emitted while satisfying all those conservation requirements.

If you just consider energy and spin conservation, the total amount of energy available in electron transitions in an atom is small, and not enough to make any of the other massive Bosons. To use your terminology, the maximum energy difference in electron transitions, Δ, is way below the energy Δ you would need to create any of the other known massive particles that satisfy the other conservation requirements.

Marco Ocram
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    I like this answer. Much clearer than accepted in what it explains. We're just not aware of particles with a mass small enough to be created by an electron transition, and neither are they predicted. (and they would probably throw all current theories of quantum physics on the scrapheap if we found any.) – Gloweye Oct 01 '19 at 07:36
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    @Gloweye PM 2Ring points out neutrinos in the comments on the question, which are probably light enough, but excluded as they state, for coupling to the weak force and not electromagnetism. – mbrig Oct 01 '19 at 22:30
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    An electron neutrino would have a sufficiently low mass, but it wouldn't satisfy the other conservation requirements, as its not a boson. – Marco Ocram Oct 01 '19 at 22:42
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    It might help to put a number on the energy of a transition and have one or two light (candidate) particle masses to see just how big the gap is. – Eric Towers Oct 02 '19 at 13:01
  • Thank you for cool explanation@marco-ocram – baponkar Oct 02 '19 at 15:00
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You are asking about when a electron transitions from a higher energy level to a lower energy level so I will assume you are asking about relaxation. There are other type of transitions, like the Auger effect.

Now when a atom/electron relaxes, it moves from a higher energy level to a lower energy level as per QM, as you state, and releases a photon.

You are asking why this transition releases a massless gauge boson.

Now the photon is an elementary particle, part of the SM, massless, pointlike. Photons always travel at speed c in vacuum, when measured locally.

You are asking why this gauge boson is massless.

There are two ways to look at this:

  1. gauge invariance, if you want to describe a theory with a zero mass vector and relativity, you have to have gauge invariance. And the photon is massless because it is the mediator of the EM force which is long range. It is because of the unbroken U(1) gauge invariance of the EM force.

Though, the gauge fields may become massive via Higgs (W,Z bosons). But that is a short range force.

Why can't gauge bosons have mass?

How does gauge invariance prevent the photon from acquiring a mass?

  1. As per SR, anything that travels at the speed of light, cannot have rest mass. This is because it would cost an infinite amount of energy to speed up a massive particle to speed c.

https://en.wikipedia.org/wiki/Photon

https://en.wikipedia.org/wiki/Special_relativity

Árpád Szendrei
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  • Yeah! I like that.Thanky you – baponkar Sep 30 '19 at 17:27
  • "unbroken U(1) gauge invariance of the EM force" = electric potential? – John Dvorak Oct 01 '19 at 13:16
  • There's one thing to keep in mind though: the photon only has to be massless ($m=0$) if it propagates further than distances $~1/m$ (in natural units). If it is immediately re-absorbed, scatters off a nearby electron and the like it may be massive (in the sense that the momentum transfer $p_{\mu}p^{\mu}$ via the photon may be different from zero). All these processes are unlikely for atoms in realistic conditions. The point is that in Quantum Field Theory only a free photon is massless (in a mass-shell renormalization scheme, don't get me started on $\gamma Z$ mixing and other wonders of QFT) – tobi_s Oct 02 '19 at 06:12
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It's just a conservation of energy equation. An example of such a case when an electron jumps to a lower shell it emits a photon and this photon itself is captured by an electron in the outer shell and that electron gets emitted from the atom. So in principle till all the conservation laws are satisfied there is always a finite but very less probability for it to happen than just a emission of photon when an electron jumps to a lower shell.

Debobrata
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