Answer:
We know that in the decay process the sum of molecular number as well as molecular weight should be constant.
The following three reaction are as follows
1 .
Alpha decay of parent nuclide
[tex]_{92}^{235}\textrm{U} \rightarrow _{90}^{231}\textrm{Th}+_{2}^{4}\textrm{alpha }[/tex]
The molecular number of alpha particle is 2 and molecular weight is 4 g/mol.
2.
Beta decay of daughter nuclide
[tex]_{90}^{231}\textrm{Th}\rightarrow _{91}^{231}\textrm{Pa}+_{-1}^{0}\textrm{beta}+v[/tex]
v is the neutrino emission,The charge on the beta particle is zero.
3.
Alpha decay
[tex]_{91}^{231}\textrm{Pa}\rightarrow_{89}^{227}\textrm{Pa}+_{2}^{4}\textrm{alpha}[/tex]
Answer:
Explanation:
The first nuclear decay that gave of alpha particle
²³⁵₉₂U → ⁴₂He + ²³¹₉₀Th
The first disintegration gave of alpha particle ⁴₂He converting the original ²³⁵U to ²³¹₉₀Th (Thorium).
This Thorium is unstable and gave off a beta particle. Beta particle are know to have a charge of an electron (e-)
²³¹Th → ⁰-₁ e + ²³¹₉₁Pa
Thorium released a beta particle to produce ²³¹₉₁Pa (protoactinum) in the process.
The protoactinum further disintegrated to produce Actinum releasing an alpha particle (⁴₂He) in the process.
²³¹Pa → ⁴₂He + ²²⁷₈₉Ac
The final product of the nuclear decay is Actinum ²²⁷₈₉Ac.
