Question

Write three nuclear equations to represent the nuclear decay sequence that begins with the alpha decay of u-235 followed by a beta decay of the daughter nuclide and then another alpha decay.

Answer 1

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.

Answer 2

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

$_{92}^{235}\textrm{U} \rightarrow _{90}^{231}\textrm{Th}+_{2}^{4}\textrm{alpha }$

The molecular number of alpha particle is 2 and molecular weight is 4 g/mol.

2.

Beta decay of daughter nuclide

$_{90}^{231}\textrm{Th}\rightarrow _{91}^{231}\textrm{Pa}+_{-1}^{0}\textrm{beta}+v$

v is the neutrino emission,The charge on the beta particle is zero.

3.

Alpha decay

$_{91}^{231}\textrm{Pa}\rightarrow_{89}^{227}\textrm{Pa}+_{2}^{4}\textrm{alpha}$