Question: What is the change in the nucleus that results from the following decay scenarios?

(a) emission of a \({\rm{\beta }}\) particle

(b) emission of a \({{\rm{\beta }}^{\rm{ + }}}\) particle

(c) capture of an electron

Beta decay (\({\rm{\beta - }}\)decay) is a type of radioactive decay in which an atomic nucleus emits a beta particle (fast, energetic electron or positron) that transforms the original nuclide into an isobar of that nuclide.

(a)

When a beta particle is emitted, a neutron is converted to a proton. This is represented as:

\(_{\rm{0}}^{\rm{1}}{\rm{n}} \to _{\rm{1}}^{\rm{1}}{\rm{p + }}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e}}\)

Therefore, the resultant equation is obtained as \(_{\rm{0}}^{\rm{1}}{\rm{n}} \to _{\rm{1}}^{\rm{1}}{\rm{p + }}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e}}\).

(b)

A proton gets converted to a neutron when there is an emission of a\({\rm{bet}}{{\rm{a}}^{\rm{ + }}}\)particle. This is represented as:

\(_{\rm{1}}^{\rm{1}}{\rm{p}} \to _{\rm{0}}^{\rm{1}}{\rm{n + }}_{{\rm{ + 1}}}^{\rm{0}}{\rm{e}}\)

Therefore, the resultant equation is obtained as \(_{\rm{1}}^{\rm{1}}{\rm{p}} \to _{\rm{0}}^{\rm{1}}{\rm{n + }}_{{\rm{ + 1}}}^{\rm{0}}{\rm{e}}\).

(c)

A proton converts to a neutron when an electron is captured. This is represented as –

\(_{\rm{1}}^{\rm{1}}{\rm{p + }}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e}} \to _{\rm{0}}^{\rm{1}}{\rm{n}}\)

Therefore, the resultant equation is obtained as\(_{\rm{1}}^{\rm{1}}{\rm{p + }}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e}} \to _{\rm{0}}^{\rm{1}}{\rm{n}}\).