Answer:
The product of the decay its Sulfur-32
Explanation:
Phosphorus-32 ( lets write it [tex]_{15}^{32}P[/tex], where the number above its the atomic mass and the number below the atomic number) decays turning a neutron into a proton and emitting radiation on the form of a electron. This is the beta minus decay, and, actually, an electronic antineutrino its also produced. We can write this decay for an X isotope with a Y isotope produced as:
[tex]_{Z}^{A}X \to _{Z+1}^{A}Y + e^- + \bar{\nu_e}[/tex]
where [tex]e^-[/tex] its the electron, and [tex]\bar{\nu_e}[/tex] the electronic antineutrino . We can see that the atomic number increases by one (cause a proton it produced and retained into the nucleus), and the atomic mass is approximately the same (there is a small difference between the neutron and proton mass, but its very small).
So, Phosphorus-32 (atomic number 15) will turn to an element with atomic number 16, and atomic mass 32, as:
[tex]_{15}^{32}P \to _{15+1}^{32}Y + e^- + \bar{\nu_e}[/tex].
[tex]_{15}^{32}P \to _{16}^{32}Y + e^- + \bar{\nu_e}[/tex].
The Y isotope must have an atomic number of 16 and an atomic mass of 32. The element with atomic number 16 its Sulfur (S), so, our decay its
[tex]_{15}^{32}P \to _{16}^{32}S + e^- + \bar{\nu_e}[/tex].
and the product of such decay its Sulfur-32
