Question

Determine the identity of the daughter nuclide from the beta decay of 8938Sr.9039Y8536Kr8734Se9038Sr8939Y

Answer 1

Answer: The daughter nuclide formed by the beta decay of given isotope is $_{39}^{89}\textrm{Y}$

Explanation:

Beta decay is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.

The released beta particle is also known as electron.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

We are given:

Parent isotope = $_{38}^{89}\textrm{Sr}$

The chemical equation for the beta decay process of $_{38}^{89}\textrm{Sr}$ follows:

$_{38}^{89}\textrm{Sr}\rightarrow _{39}^{89}\textrm{Y}+_{-1}^0\beta$

Hence, the daughter nuclide formed by the beta decay of given isotope is $_{39}^{89}\textrm{Y}$