I'm not sure how I can determine whether an element has $\beta^-$ or $\beta^+$ emission, or no $\beta$ emission at all.

I'm told that positron decay happens when there are too many protons and not enough neutrons, but what does 'too many' mean? For example, I know $^{15}O$ has 8 protons and 7 neutrons and it has $\beta^+$ decay, but why is the neutron not enough? Does $^{14}O$ have beta decay?

In which situations does an isotope have $\beta^-$ decay, or no beta decay at all?


3 Answers 3


How to determine if an element undergoes electron or positron emission?

This is actually a false choice!

For example 40K decays in three ways:

  • $\beta^-$ 89.28% (emits $e^- + \bar{\nu}$)
  • $\beta^+$ 0.001% (emits $e^+ + \nu$)
  • EC 10.72% (captures an atomic electron, emits $\nu$ plus usually some accompanying atomic photons)

So nuclear wallet cards while incredibly handy aren't the last word in nuclear physics.

update!: @PM2Ring's comment allerts us of even more nuclear ambivalency or creativity:

Some isotopes can do both + and - beta decay, eg Cu-64. Also see K-40

From Wikipedia's Potassium-40:

Potassium-40 is a rare example of an isotope that undergoes both types of beta decay. In about 89.28% of events, it decays to calcium-40 (40Ca) with emission of a beta particle (β−, an electron) with a maximum energy of 1.31 MeV and an antineutrino. In about 10.72% of events, it decays to argon-40 (40Ar) by electron capture (EC), with the emission of a neutrino and then a 1.460 MeV gamma ray. The radioactive decay of this particular isotope explains the large abundance of argon (nearly 1%) in the Earth's atmosphere, as well as prevalence of 40Ar over other isotopes. Very rarely (0.001% of events), it decays to 40Ar by emitting a positron (β+) and a neutrino

It's hard to get the branching ratios exactly right because the decay by three different modes, some radiative, some to the ground state require different techniques to measure, and their efficiencies are difficult to normalize accurately.

Nonetheless here's a decay diagram from Hyperphysics that cites some historic work. Similar can be found in The K/Ar dating method : principle, analytical techniques, and application to Holocene volcanic eruptions in Southern Italy:

Even though the decay of 40K is somewhat complex with the decay to 40Ca and three pathways to 40Ar, Dalrymple and Lanphere1 point out that potassium-argon dating was being used to address significant geological problems by the mid 1950's. The energy-level diagram below is based on data accumulated by McDougall and Harrison2.

1Dalrymple, G. Brent and Lanphere, Marvin A., Potassium-Argon Dating, W.H. Freeman, 1969.

2.McDougall, Ian and Harrison, T. Mark, Geochronology and Thermochronology by the 40Ar/39Ar Method, 2nd Ed., Oxford, 1999.

"The energy-level diagram below is based on data accumulated by McDougall and Harrison" from http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/KAr.html

  • 3
    $\begingroup$ You’re correct that “nuclear wallet cards aren’t the last word,” but your alternative is … Wikipedia? Wikipedia is great, but only when it’s great. For example, a year ago, a helpful person replaced the detailed decay probabilities in the copper-64 page with different values from an outdated (2004) reference, but did not fix the misstatement that copper-64 decays by gamma radiation. (The description of the gamma ray in your quoted paragraph about potassium is correct.) $\endgroup$
    – rob
    Feb 20, 2022 at 12:50
  • $\begingroup$ @rob I forget how nice people talk to each other here in Physics SE, so much better than it was before. Are you complaining that I found a source that supplements the other source and seems more complete for this nuclide, or that I advocated that we shouldn't trust single non-primary sources, or that I called going both ways "nuclear ambivalency or creativity" or that I cited PM2Ring's comment and gave them a hat tip? Or is it that I failed to cite paywalled sources in order to look smart? Why the harsh words exactly; What open source for potassium 40 decay branching ratios would you suggest? $\endgroup$
    – uhoh
    Feb 20, 2022 at 14:43
  • $\begingroup$ @rob moderators can set the tone, and as one, going after me rather than making a helpful, friendly comment suggesting ways to improve the post tells other readers that that's a good way to use comments. Nonetheless I've added something from Hyperphsyics. $\endgroup$
    – uhoh
    Feb 20, 2022 at 14:50
  • 5
    $\begingroup$ Apologies if my frustration seemed to be directed at you. It regularly happens that a helpful Wikipedia editor will replace a result from a systematic global literature review with a result from a single experiment or (in this case) a result which has been superseded. I linked elsewhere to the NNDC, which has a number of databases in addition to the wallet cards, and excellent citation hygiene. But for this question that's like drinking from the firehose. $\endgroup$
    – rob
    Feb 20, 2022 at 15:59
  • $\begingroup$ @rob thanks! I have found that simply stopping whenever I write the word "you" and seeing if I can rephrase to focus on improving a post or addressing an issue rather than focusing on an individual helps me to write happier, more actionable comments. $\endgroup$
    – uhoh
    Feb 20, 2022 at 16:40

You can check the Nuclear Wallet Cards, hosted by the National Nuclear Data Center, for the isotope you have a question about.

For example, if we look at all of the data for mass number $A=14$,

data table

then we see that nitrogen-14 is the only stable nuclide with this mass number. Carbon-14 decays by $\beta^-$ emission. The decay mode "$\epsilon$" for oxygen-14 means "$\beta^+$ decay mixed with electron capture." Apparently fluorine-14 is already the proton drip line.

As to why these decays are allowed: nitrogen is the beta-decay endpoint for $A=14$ because it is the isotope with that mass number but the smallest actual mass (tabulated here, in the column $\Delta$, as the mass excess).

  • 6
    $\begingroup$ @IGY Some isotopes can do both + and - beta decay, eg Cu-64. Also see K-40 $\endgroup$
    – PM 2Ring
    Feb 20, 2022 at 4:08
  • $\begingroup$ @PM2Ring good point! I've cited you here $\endgroup$
    – uhoh
    Feb 20, 2022 at 6:21

The reason is the strong nuclear force. This is the most powerful force in nature (within suitable conditions). This is the force that glues protons and neutrons together. This force is stronger than the electrostatic force. This force has some properties which I will not specify, but you can read What is Nuclear Force? – Definition, Properties, Examples.

But as the number of protons in the nucleus increases the electrostatic force starts dominating over the strong nuclear force. To avoid this, you will see a pattern as the number of protons in a nucleus increases. This graph:

Enter image description here

You can see that there is a exponential increase in the number of neutron and we increase the atomic number (Z). The strong nuclear force doesn't depend upon the charge so as the number of protons increase and the electrostatic force starts dominating, and the nucleus needs more neutrons to stabilise. That's were the weak nuclear force comes into play (which we will not discuss). To stabilise the nucleus, the protons starts converting into neutrons, positrons and a neutrino (which is 𝛽+ decay).

For 14O, it is a hybrid of 𝛽+ and electron capture. If you want to find what the ejected fermion (particles with 1/2 spin +or-) is, in this case an electron or positron, it is through using a magnetic or electric field (further is only calculation through which you can find the particle).

  • $\begingroup$ The GeeksforGeeks article doesn't leave out too many definite and indefinite articles (very often the case on GeeksforGeeks) - though it does miss a definite article in the title. But it still has a lot of weird capitalisation of words. $\endgroup$ Feb 21, 2022 at 15:38
  • $\begingroup$ (For emphasis and other formatting, we have italics, bold, superscript², subscript, headlines, and MathJax on this platform - though it shouldn't be overused.) $\endgroup$ Feb 21, 2022 at 16:00
  • $\begingroup$ What do you mean by "further is only calculation through which you can find the particle" (seems incomprehensible)? $\endgroup$ Feb 21, 2022 at 16:00
  • $\begingroup$ Further calculation means passing that particle through a electric or magnetic field and finding out the direction of deflection. After you get direction of deflection you can find the magnitude of charge using right hand rule(in presence of magnetic field) and in electric field you can find out using the formula(I will not elaborate it.). $\endgroup$ Feb 25, 2022 at 16:18
  • $\begingroup$ Can you update your answer, please (but without "Edit:", "Update:", or similar - the answer should appear as if it was written today). $\endgroup$ Jun 13, 2022 at 15:47

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