# Why does the product nucleus lose an electron during beta positive decay process?

The question describes the beta positive decay for Nitrogen-13 into Carbon-13. After reading the explanation for the answer, it says that carbon ion will lose an electron in the decay process; therefore the the mass of products is the carbon nucleus + electron + positron.

But I'm just confused because the equation does not include an electron, yet its mass is added to the mass of the products. Why is that? and why isn't it the same for beta negative decay?

Reaction provided in the question: Nitrogen-13 -> Carbon-13 + Positron + neutrino

Atomic Masses:

Nitrogen-13: 13.005739 u

Carbon-13: 13.003355 u

Electron/Positron: 0.000549 u

What I did:

E=Δmc^2

Δm=m(parent)-m(products)

Δm=nitrogen-13-(Carbon-13 + positron mass)

Δm= (13.005739u)-(13.003355u+0.000549u)

Δm= 0.001835 (931.5MeV/u)

E= 1.709 MeV

Solution:

E=Δmc^2

Δm=m(parent)-m(products)

Δm=nitrogen-13-(Carbon-13 + electron mass + positron mass)

Δm= (13.005739u)-(13.003355u+ 2(0.000549u))

Δm= 0.001286 (931.5MeV/u)

E= 1.198 MeV

You are modeling the following reaction:

Nitrogen-13 -> Carbon-13 + Positron + neutrino

This reaction is in the nucleus, so you should use the mass of the nucleus in the mass balance, not the atomic mass. However, we generally don't have the nucleus mass available, so we use the atomic masses. The atomic masses include the mass of the electrons.

The equation you should be using:

Δm=nucleus(N13)-(nucleus(C13) + positron mass)

Add 7 electrons to each side to use atomic mass:

Δm=nucleus(N13)+ 7(electron mass) -(nucleus(C13) + 7(electron mass) + positron mass)

Combine the nucleus mass and electron mass to get atomic mass. Note that the carbon atom only has 6 electrons, so you are left with an extra electron.

Δm=atomic(N13) -(atomic(C13) + electron mass + positron mass)

This is where the extra electron mass comes from.

A $$β^+$$ decay goes with the following diagram:

The leading-order Feynman diagram for $$β^+$$ decay of a proton into a neutron, positron, and electron neutrino via an intermediate $$W^+$$ boson

In other words a proton becomes a neutron, the charge of the proton leaving with the positron. This means that the positive charge of the nucleus diminishes by 1, Z goes to Z-1, and the atom is negatively ionized. This extra electron left over from the disappearance of the proton, has to be taken into account into energy balances for the decay.

It is not the nucleus that loses an electron, the electron is left over from the atomic levels of the decaying nucleus.

I found this:

The atomic mass is a weighted average of all of the isotopes of that element, in which the mass of each isotope is multiplied by the abundance of that particular isotope

So it depends whether the masses you are given as atomic mass are as the above definition for the periodic table, or it is the mass of the whole atom. If it is the mass of the whole atom, the ionization electron has to be taken in account in the energy balance.

• I am always sad when I get a down vote without the reason why, to allow me to improve . Dec 21, 2021 at 7:27
• The Atomic Mass Evaluation 2020 is readily available, say through www-nds.iaea.org/amdc - then you have the latest assessment of the mass of each nucleus. Dec 21, 2021 at 14:35
• @JonCuster a file talsk of "rounded masses" in that case the electron is irrelevant. The OP is asking of relevance of the electron, so it must be the atomic mass including the electrons in the problem. Dec 21, 2021 at 15:54
• I believe the definition of the atomic mass you are using is the average atomic mass of an element. Usually the atomic mass is defined for a single isotope. Dec 24, 2021 at 14:30
• @NuclearFission I am not using it, I am pointing out ( with links) there are different definitions and the answer will depend on the definition of atomic mass. Dec 24, 2021 at 18:12