# In nuclear chemistry, how does a neutron split to form a proton and an electron?

I'm studying radioisotopes at the moment and balancing nuclear reactions isn't making sense in that more matter is coming out of the equation in negative β decay equations :

$^\textrm{14}_\textrm{6}\textrm{C}$ → $^\textrm{14}_\textrm{7}\textrm{N}$ + $\textrm{e}^\textrm{-}$ + $\textrm{v}^\textrm{-}_\textrm{e}$

Notice how the original element has decayed into a new element with an unchanged mass number but an atomic number that has increased by one.

In positive β decay equations, it makes sense:

$^\textrm{23}_\textrm{12}\textrm{Mg}$ → $^\textrm{23}_\textrm{11}\textrm{Na}$ + $\textrm{e}^\textrm{+}$ + $\textrm{v}^\textrm{e}$

How can you create something with the same mass but with another proton?

• $n \rightarrow p + e^- + \bar{\nu}_e$ – user26143 Dec 1 '13 at 22:36

One neutron has changed into a proton, that's what has happened. Our convention in chemistry is to identify nuclear species by the proton number (or the "atomic number") and that is why a new nuclide with an increased atomic number is formed. Both the $\beta^+$ and $\beta^-$ decay follow identical paths: in $\beta^+$ decay, a neutron changes into a proton (thus giving no change in mass number and increase in atomic number), whereas in $\beta^-$ decay, a proton decays into a neutron giving no change in the mass number and a decrease in the atomic number.