Determine the identity of the daughter nuclide from the beta decay of $\begin{array} { l }
32 \
15
\end{array}$ P.

Question

Determine the identity of the daughter nuclide from the beta decay of $\begin{array} { l } 
32 \
15
\end{array}$ P.

Quizplus · Accepted Answer

Beta decay is the process in which a neutron in the nucleus of an atom spontaneously transforms into a proton, accompanied by the emission of an electron (beta particle) and an antineutrino. This process results in the increase of the atomic number by 1 and the mass number remains the same. Therefore, the daughter nuclide will have the same mass number as the parent nuclide but its atomic number will be increased by one.

In this problem, the parent nuclide is $\begin{array} { l } 
32 \
15
\end{array}$ P, which has a mass number of 32 and an atomic number of 15. After beta decay, the atomic number will increase by one and become 16, while the mass number remains 32. The only choice that fits this description is B, which is $\begin{array} { l } 
32 \
16
\end{array}$ S (sulfur).

Determine the Identity of the Daughter Nuclide from the Beta 3215\begin{array} { l } 32 \\15\end{array}3215​

Determine the Identity of the Daughter Nuclide from the Beta $\begin{array} { l } 32 \\15\end{array}$