Question

How does an atom with too many neutrons relative to protons undergo radioactive decay?

Answer 1

We know that if an atom consists of a high number of neutrons with respect to protons then it will be very unstable and it undergoes $\beta -decay$. This means that the atom stabilizes itself by emitting $\beta -particles$ from the atomic nucleus, transforming the original nuclide to an isobar of that nuclide.

Complete step by step answer:
Let us first discuss about $\beta -decay$ and $\beta -particles$:-
Beta particles are electrons or positrons (electrons with positive electric charge are referred as positrons) which are emitted during radioactive decay. Beta decay occurs when a nucleus has either too many protons or too many neutrons due to which it becomes an unstable nuclei. Then one of the protons or neutrons is transformed into the other and this leads to the formation of beta particles. In beta minus decay (${{\beta }^{-}}$ ), a neutron decays into a proton, an electron, and an antineutrino whereas in case of beta plus decay (${{\beta }^{+}}$), a proton decays into a neutron, a positron, and a neutrino. These particles are very fast moving electrons.
- As it is mentioned that number of neutrons is much greater than number of protons, therefore neutrons will get transformed into protons, electrons and an antineutrino i.e., beta minus decay (${{\beta }^{-}}$ ) will take place.
The example of beta minus decay (${{\beta }^{-}}$ ) is shown below:-
$_{Z}^{A}X\to _{Z+1}^{A}X+_{-1}^{0}\beta +\overline{{{v}_{e}}}$
We can see that a neutron got transformed into proton (Z+1) and electron (-1).
- This way an atom stabilizes itself by undergoing $\beta -decay$.

Note: Electron capture is also included as third type of beta decay, because the basic nuclear process mediated by the weak force is almost the same. In this an inner atomic electron is captured by a proton in the nucleus and transforms it into a neutron and also releases electron neutrinos.