How can I determine if an element undergoes electron or positron emission?

Question

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?

Answer 1

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

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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

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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.

Answer 2

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 $A=14$,

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).

Answer 3

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:

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).