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

${}_{Z-1}^{A}Y$

In a Beta+ decay, a proton in the parent nucleus decays into a neutron that remains in the daughter nucleus and the nucleus emits a positron and a neutrino. 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:
$\begin{array}{c}{}_Z^{A}X{\to }_{Z-1}^{A}Y+e^{+}+v\end{array}$