Respuesta :
Answer: The 4 nuclear equations for the given series of emissions are written below.
Explanation:
Alpha decay is the process in which the nucleus of an atom disintegrates into two particles. The first one which is the alpha particle which consists of two protons and two neutrons, also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.
In this decay process, the atomic number of the parent nucleus decreases by a factor of 2 and the mass number also decreases by a factor of 4
[tex]_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha[/tex]
Beta decay is defined as the process in which the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.
[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]
The chemical equation for the alpha decay of Th-232 isotope follows:
[tex]_{90}^{232}\textrm{Th}\rightarrow _{88}^{228}\textrm{Ra}+_2^4\alpha[/tex]
The chemical equation for the beta decay of Ra-228 isotope follows:
[tex]_{88}^{228}\textrm{Ra}\rightarrow _{89}^{228}\textrm{Ac}+_{-1}^0\beta[/tex]
The chemical equation for the beta decay of Ac-228 isotope follows:
[tex]_{89}^{228}\textrm{Ac}\rightarrow _{90}^{228}\textrm{Th}+_{-1}^0\beta[/tex]
The chemical equation for the alpha decay of Th-228 isotope follows:
[tex]_{90}^{228}\textrm{Th}\rightarrow _{88}^{224}\textrm{Ra}+_2^4\alpha[/tex]
Hence, the 4 nuclear equations for the given series of emissions are written above.
