# How to determine if this is β- or β+ decay?

## Homework Statement

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The problem states that 24 11Na is radioactive. The question asks if it's a β- or β+ emitter. The 24 is the atomic mass number and the 11 is the atomic number.

## Homework Equations

I know that in β- decay, the atomic number increases by 1 and it emits an electron and antineutrino.
For β+ decay, the atomic number decreases by 1 and it emits a positron and neutrino.

## The Attempt at a Solution

I know that 24-11 = 13 neutrons. So there are more neutrons than protons (11). How can I use this information to figure out if it will undergo β- or β+ decay?

haruspex
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What are the decay products for each case?

Orodruin
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The rule of thumb for light nuclei is that they decay toward a more even distribution of protons and neutrons. This is also true here.

For heavier nuclei this is no longer the case. The stability line then corresponds to a larger number of neutrons.

What are the decay products for each case?
If it undergoes β- then it would look like this: 24 11Na -> 24 12 Mg + β- + antineutrino
If it undergoes β+ then it would look like this: 21 11Na -> 24 10 Ne + β+ + neutrino

Knowing this, I'm not sure how to proceed.

haruspex
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If it undergoes β- then it would look like this: 24 11Na -> 24 12 Mg + β- + antineutrino
If it undergoes β+ then it would look like this: 21 11Na -> 24 10 Ne + β+ + neutrino

Knowing this, I'm not sure how to proceed.
I was going to suggest that it would be the one with the greater net energy release, but Orodruin likely has more reliable information.

The rule of thumb for light nuclei is that they decay toward a more even distribution of protons and neutrons. This is also true here.

For heavier nuclei this is no longer the case. The stability line then corresponds to a larger number of neutrons.
I'm not quite sure what you mean by this. What I'm understanding is that because there are more neutrons than protons, then the nucleus becomes unstable. A stable nucleus has an even amount of neutrons and protons. So because there are more neutrons than protons, the neutron must change to a proton. I'm not sure I'm thinking about this correctly.

Orodruin
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I was going to suggest that it would be the one with the greater net energy release, but Orodruin likely has more reliable information.

Typically beta unstable nuclei only have one possible direction, the other would be energetically forbidden. However, there are some rare instances of odd-odd nuclei that actually have both channels open. An example is 74As that has a branching ratio of 2/3 of ##\beta^+## decay into 74Ge and a branching ratio of 1/3 of ##\beta^-## to 74Se.

I'm not quite sure what you mean by this. What I'm understanding is that because there are more neutrons than protons, then the nucleus becomes unstable. A stable nucleus has an even amount of neutrons and protons. So because there are more neutrons than protons, the neutron must change to a proton. I'm not sure I'm thinking about this correctly.
Not necessarily. As I said, it is a rule of thumb, nothing more. It is also only applicable to light nuclei. See https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html
Note that even for the lighter nuclei, stable odd-odd nuclei are very rare and the stable nucleus with the same atomic number tends to have 2 more neutrons than protons when it is not possible to have the same number of protons and neutrons. The light stable odd-even nuclei tend to have one more neutron than protons.

haruspex
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Typically beta unstable nuclei only have one possible direction, the other would be energetically forbidden.
So that does fit with my supposition.
there are some rare instances of odd-odd nuclei that actually have both channels open. An example is 74As that has a branching ratio of 2/3 of ##\beta^+## decay into 74Ge and a branching ratio of 1/3 of ##\beta^-## to 74Se.
Do those also reflect relative energy releases?

Orodruin
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Do those also reflect relative energy releases?
Yes, but I would advise against taking it as a completely general supposition. The Q value of the ##\beta^+## is 2.6 keV and that of the ##\beta^-## is 1.4 keV. You can find all of this information in the table of nuclides I linked to in #7.

Edit: Just as a counter example, the isotope 112In has almost equal beta+ and beta- branching ratios. The Q value for the beta+ is roughly 5 times larger than that for the beta-.

• haruspex
haruspex