2
$\begingroup$

As we know that $\beta^-$ decay is a nuclear decay which is only concerned with the nucleus, then how come a parent nuclei can emit an electron which is outside the nucleus in $\beta^-$ decay, what exactly happens inside the nucleus for the emission of an electron and also an Anti-neutrino?

$\endgroup$
10
  • 1
    $\begingroup$ See beta decay, the $\beta^{-}$ section. $\endgroup$ Dec 19, 2020 at 9:42
  • $\begingroup$ I have seen that but it does not make my concept clear, it only says that during the decay these particles are emitted but do not mention the logic behind that. $\endgroup$
    – Autodidact
    Dec 19, 2020 at 10:16
  • 2
    $\begingroup$ Apart from going into the details of the Glashow-Weinberg-Salam model, we cannot say anything but "this is the way nature behaves like". $\endgroup$ Dec 19, 2020 at 10:44
  • 1
    $\begingroup$ What does "a parent nuclei can emit an electron which is outside the nucleus" mean? $\endgroup$
    – PM 2Ring
    Dec 19, 2020 at 11:20
  • $\begingroup$ It was used for electron. $\endgroup$
    – Autodidact
    Dec 19, 2020 at 12:29

2 Answers 2

2
$\begingroup$

Nuclides ( the nuclei of atoms) are either stable , i.e. never decay, or unstable and have a probability of decay.

nuclides

Graph of nuclides (isotopes) by type of decay. Orange and blue nuclides are unstable, with the black squares between these regions representing stable nuclides. The continuous line passing below most of the nuclides comprises the positions on the graph of the (mostly hypothetical) nuclides for which proton number would the same as neutron number. The graph reflects the fact that elements with more than 20 protons either have more neutrons than protons or are unstable.

Various quantum mechanical models exist to describe nuclei, and they all depend on parametrising the fact that in order to have a stable nucleus there must be a large number of neutrons to counteract, by the strong nuclear interaction, the repulsive effect of the proton charges.

In particular for beta decay, the fact that a free neutron decays via the weak interaction, explains the beta decay for nuclides with a high number of neutrons, because there will be a quantum mechanical probability for a core neutron to be far enough and long enough away from the center of mass for the neutron to decay via the weak interaction. Energy conservation should also apply:

Because the reaction will proceed only when the Q value is positive, β− decay can occur when the mass of nucleus X of given Z( number of protons) is greater than the mass of nucleus X' with Z+1 protons .

$\endgroup$
1
  • $\begingroup$ Thank you for answering this question and your deep explanation. $\endgroup$
    – Autodidact
    Dec 19, 2020 at 15:46
0
$\begingroup$

I think $\beta$-decay happens because there are neutrons in higher Energy-Levels than protons (or the other way around), and it is therefor favorable for a neutron to turn into a proton. For example the Boron-12 nucleus has two neutrons more than protons, whereas the carbon-12 nucleus has as many neutrons as protons. More specifically, one of the Boron-12's neutrons is in a higher Energy-Level than the the highest Energy-Level of the protons and there is a spot for a proton at that Energy level. Therefor $\beta$-decay occures: $$ {}^{12}_5B\longrightarrow {}^{12}_6C+e^-+\nu^*. $$

$\endgroup$
1
  • $\begingroup$ Thank you for answering this question $\endgroup$
    – Autodidact
    Dec 19, 2020 at 15:46

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.