Chemistry: Determine the identity of the daughter nuclide from the beta decay of 8938sr.9039y8536kr8734se9038sr8939y

Determine the identity of the daughter nuclide

imhorribleatmath90

12.01.2022

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 $_{39}^{89}\textrm{Y}$

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jazminpratt0311

04.10.2022

Green.

Explanation:

Only the particles that hit the nucleus of the atom bounced off.

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