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Beta Minus Decay

by omiros
Tags: beta, decay, electron, minus, nucleus
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omiros
#1
Jun19-13, 09:54 AM
P: 30
Hello everybody, I am a first year physics student and I have a question about Nuclear Beta Minus Decay.

I was thinking the other day, about a beta decay. After the nucleus is formed, the new atoms state is a positive ion with charge +1.

If we think of the electron escaping from somewhere close to the nucleus the electron will be pulled by the nucleus.

Is that electron bound at any time at all, or not?

I understand that the function that describes the acceleration of the electron is going to be very weird, but I just care about the final kinetic energy of both.

Also what happens to the rest of the electrons? How are they going to react in such a case? Do any of they emit photons? Do we usually have collision between that electron and the 'bound' electrons?
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Bill_K
#2
Jun19-13, 12:08 PM
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Quote Quote by omiros View Post
If we think of the electron escaping from somewhere close to the nucleus the electron will be pulled by the nucleus. Is that electron bound at any time at all, or not? I understand that the function that describes the acceleration of the electron is going to be very weird, but I just care about the final kinetic energy of both.
Atomic electron energy levels are rather small compared to the energies involved in beta decay. So the emitted electron would not be bound, although it's true it will be a bit slowed down escaping the atom, and this correction needs to be taken into account when observing the decay's energy spectrum.

Also what happens to the rest of the electrons? How are they going to react in such a case? Do any of they emit photons? Do we usually have collision between that electron and the 'bound' electrons?
It's possible, although infrequent, for the emitted electron to collide with an atomic electron. A more interesting example of the interplay between nuclear decay and the atomic electrons is an alternative decay mode to beta plus decay called electron capture or K-capture, in which the nucleus grabs an atomic electron. Since this electron is taken from a low-lying shell, the atom needs to fill the hole, by emitting an X-ray, or sometimes a second ("Auger") electron.
snorkack
#3
Jun20-13, 02:44 AM
P: 390
Quote Quote by Bill_K View Post
Atomic electron energy levels are rather small compared to the energies involved in beta decay.
Most of time. Not always.
The decay energy of rhenium 187 is just 2,6 keV, whereas the binding energy of the inner electrons of heavy atoms is in hundreds of keV

The rhenium 187 nucleus is over 1 milliard times shorter lived than the neutral atoms. It follows that when a rhenium 187 nucleus undergoes beta decay, over 99,999999% times the electron is not emitted but goes into some bound state (ground or excited).
Neutral dysprosium 163 atom is completely stable, so the electron is always bound.

The beta decay energy is randomly divided between electron and antineutrino. Even if atomic energy levels are small, there is small but nonzero chance that the antineutrino happens to get almost all beta decay energy and the electron gets little enough to stay bound to the atom.
Quote Quote by Bill_K View Post
A more interesting example of the interplay between nuclear decay and the atomic electrons is an alternative decay mode to beta plus decay called electron capture or K-capture, in which the nucleus grabs an atomic electron. Since this electron is taken from a low-lying shell, the atom needs to fill the hole, by emitting an X-ray, or sometimes a second ("Auger") electron.
K-capture depends on the choice of shell whence the electron is captured. Only s electrons can ever be captured, because only they go to nucleus - but all s electrons do, not just the 1s ones. The probability of K-capture is simply bigger than L-capture or higher shell captures... except when K-capture is impossible. As is the case with holmium 163.


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