Question

A nucleus ${}_Z^{A}X$ going through a ${\beta }^{+}$ decay will end up taking what configuration?

• A. ${}_{Z+1}^{A}Y$
• B. ${}_Z^{A-1}X$
• C. ${}_{Z+1}^{A+1}Y$
• D. ${}_{Z-1}^{A}Y.$

Answer 1

The correct option is D ${}_{Z-1}^{A}Y.$

Grasp this: in any nuclear process, a positron getting released (from a proton) is equivalent to an electron getting absorbed (by a proton).
(The following is for your understanding; although it's technically correct, do not write this when you answer subjective questions in exams, unless you understand this well enough to defend yourself. We can write the equivalence:
$[{}_Z^{A}X\to [....]+e^{+}+v]\equiv [{}_Z^{A}X\to [....]-e^{-}+v]\equiv [{}_Z^{A}X+e^{-}\to [....]+v]$.)
Visualizing, a proton from a ${}_Z^{A}X$ nucleus releasing a positron is equivalent to a proton from the ${}_Z^{A}X$ nucleus absorbing an electron.
A proton in the ${}_Z^{A}X$ nucleus absorbs an electron (equivalently, releasing a positron) to become a neutron, releasing a neutrino in the process. Thus the atomic number Z is bound to decrease by 1, but the number of nucleons being constant, the mass number A will stay the same. Thus the complete decay would be:
$\begin{array}{c}{}_Z^{A}X{\to }_{Z-1}^{A}Y+e^{+}+v\end{array}.$