# Bremsstrahlung vs energy conservation

From Wikipedia:

Bremsstrahlung is electromagnetic radiation produced by the deceleration of a charged particle when deflected by another charged particle, typically an electron by an atomic nucleus. The moving particle loses kinetic energy, which is converted into a photon because energy is conserved.

Isn't energy conserved for the moving particle in an electrostatic potential, $E_{kinetic} + E_{potential} = \frac{mv^2}{2}+\frac{kqQ}{r}$? If so, where does the extra energy for photons come from?

Why don't electrons in atoms radiate away their energy?

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"Isnt energy conserved for the moving particle, kinetic + potential, if so, where does the extra energy for photons come from?": "The moving particle loses kinetic energy, which is converted into a photon because energy is conserved." You have both asked and answered this question. –  Michael Brown Apr 17 '13 at 23:15
Hi errr: Do you have log-in problems? It might help to register your account. –  Qmechanic Apr 17 '13 at 23:41

## 1 Answer

The short answer to your question is "because of quantum mechanics". If you view the atom and the electron orbit around the nucleus as a classical system then indeed the system would not be stable and the electron would lose orbital energy. This is why Niels Bohr proposed a model with quantized energy levels.

This was one of the first serious motivations for quantum mechanics.

If you want an intuitive model to understand this, instead of thinking of the electron as some point orbiting, think of a 3D standing wave. A vibrating water drop or other shell-like membrane around the nucleus vibrating without any friction. This model isn't any more correct than the single orbiting electron but at least then you can see it not losing energy.

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Downvoter, please leave a comment when you downvote. –  Brandon Enright Apr 18 '13 at 15:45
I didn't downvote but it might be because this only answers the very last part of OP's question, i.e. why are atoms stable, which was only added in a second edit. The main question is about Brehmsstrahlung which is a separate issue. –  Mark Mitchison Apr 18 '13 at 17:18