# Clarification on isotope notation

I am a bit confused about the idea of isotope notation. I know that the top number is the mass number, equivalent to the number of neutrons + protons, while the bottom is the atomic number, equivalent to the number of protons.

What I don't understand is why during beta decay, the electron emitted is represented as having an atomic number of -1. What exactly does it mean to have a negative atomic number? Can the atomic number be considered as just the charge?

Also, my understanding is that the -1 charge balances the gain in the charge due to the proton being formed from a neutron. However, where does this electron originate from if the nucleus originally only contains protons and neutrons?

Thanks for any help, I appreciate it.

Can the atomic number be considered as just the charge?

It's not standard to talk about an electron as having an atomic number. However, if you interpret these numbers as charges, then the system you've been taught makes sense.

However, where does this electron originate from if the nucleus originally only contains protons and electrons?

Processes like these can create and destroy particles.

It is just a matter of balancing the books and counting charge.

Consider the beta decay of Carbon-14.

$$\rm ^{14}_{\,\,6}C \to {}^{14}_{\,\,7}N^+ + {}^{\;\;0}_{-1}\beta^- + {}^{0}_{0}\bar \nu$$.

$$(+6)=(+7)+(-1)$$ balances the charge.

If you do the counting of particles this decay is equivalent to

$$\rm ^{1}_{0}n \to {}^{1}_{1}p^+ + {}^{\;\;0}_{-1}\beta^- + {}^{0}_{0}\bar \nu$$

ie a neutron decaying into a proton . . . .

• *Carbon-14..... – Jasper Mar 28 at 22:09
• @Jasper Many thanks. I have corrected my mistake. – Farcher Mar 28 at 22:25

However, where does this electron originate from if the nucleus originally only contains protons and electrons?

I assume you meant protons and neutrons. As Ben Crowell already said - the electron (and the accompanying anti-neutrino) are created. The force responsible for that is the weak force. The constituent quarks of the proton are (uud) and of the neutron - (udd) (where u=up, d=down, with charges +$$\frac{2}{3}$$e and -$$\frac{1}{3}$$e respectively). The interaction that generates the leptons and turns the neutron into a proton is: $$d\rightarrow u + e^- +\overline{\nu}_e$$. A Feynman diagram of the process can be found easily on the internet, e.g. this one from Wikipedia.