# How does inner Bremsstrahlung work?

I'm trying to understand inner Bremsstrahlung. I know this applies to beta minus decay, but have a hard time understanding how it works. In the beta decay, electron is emitted from nucleus. I believe it's quite energetic/fast, but where does the electron go?

• does it completely go out of the emitted atom?
• or does it stay at the orbital shell?
• what type of photons could it emit? only X-rays?

I believe once it's emitted from the nucleus, (right away, it experiences the attraction force in the nucleus) and it should slow it down and photon might be emitted.

I can't see the whole picture, would appreciate the easier explanation than in the books.

• en.wikipedia.org/wiki/… May 2, 2023 at 10:57
• anna, well, I know how to find information on the internet, it's just I couldn't understand it fully. If only we understood everything from wikipedia, there would be no need for stackexchange to exist at all ever :)
– Nika
May 2, 2023 at 15:09

By definition, inner bremsstrahlung is arises from the creation of the electron and its loss of energy (due to the strong electric field in the region of the nucleus undergoing decay) as it leaves the nucleus (https://en.wikipedia.org/wiki/Bremsstrahlung). Otherwise it is not "inner" bremsstrahlung.

Furthermore, as any bremsstrahlung process, the emitted photons carry energy that is proportional to the loss of energy experienced by the accelerated electron. There is no infrared/ultraviolet cutoff dictated by the theory as far as I am aware of, other than the cutoff imposed by the energy conservation principle. Namely,

$$E_i=E_f+\hbar\omega\Rightarrow\hbar\omega=E_i-E_f$$

with $$E_i$$ and $$E_f$$ being the initial and final energies carried by the electron, whereas $$\omega$$ is the frequency of the emitted photon.

Extra piece of information that might turn out to be useful: If that frequency is low enough, such that the photon can not be detected by experimental devices (because of their finite resolution), we call such photons soft.

I hope this covers you. If not, you can always comment.

• So in beta minus decay, electron is emitted from the nucleus and before it leaves the nucleus, the attraction force is happening and electron’s speed decreases and photon is emitted. Thats all clear. Then electron wont be able to stay at any shell as there is no place for it due to original electrons holding the shells. So electron will leave the atom completely right ? or there is a chance that emitted electron might hit one of the original electron and displace it ? Thats where I am most interested. Thank you
– Nika
May 2, 2023 at 14:50
• @GiorgiLagidze "is a chance that emitted electron might hit one of the original electron and displace it" To displace an atomic electron from its orbital the photon energy must be equal or larger than the orbital energy. This will need a particular study of the energy possibilities of the atoms in the system. May 2, 2023 at 15:58
• I didn't mean the photon. I meant emitted electron hitting one of the original electrons. The emitted electron could still have enough binding energy even after having been decelerated by the nucleus.
– Nika
May 2, 2023 at 16:02
• This is what happens in one of the cases of photoelectric effect. Electron coming from outside the atom hits the electron, moves it out of the atom and the electron(that hit original one) reflects and go in different direction. I wonder the same thing, but in beta minus decay
– Nika
May 2, 2023 at 16:04
• @GiorgiLagidze the same energy consideration for a beta decay electron interacting with the atomic orbitals wiil hold, it is a quantum mechanical setup, one has to calculate for the particular one whether the beta electron has enough energy to change an orbital of the atom so as to eject a second electron May 2, 2023 at 18:04