Question

: β^- decay producing⁹⁰Y. The parent nuclide is a major waste product of reactors and has chemistry similar to calcium, so that it is concentrated in bones if ingested⁹⁰Yis also radioactive.

Short answer

The equation is obtained as: \[_{38}^{90}S{r_{52}} \to _{39}^{90}{Y_{51}}{ + _{ - 1}}\beta + {\bar v_e}\].

Step-by-step solution

Define Radioactivity

The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.

Explanation

The term β^- decay" refers to the emission of a beta particle. The beta particle has an atomic mass of 0 and an atomic number of - 1. As a result, the general equation may be expressed as follows:

\[_Z^AX \to _{z + 1}^A{Y_{z - 1}}{ + _{ - 1}}\beta + {\bar v_e}\]

The atomic number of ⁹⁰Y is 84.

As a result, the parent nucleus will have an atomic number of 39 - 1 = 38 before beta decay. The atomic mass of ⁹⁰Y will be 90.

The parent nucleus will have an atomic mass of 90 + 0 = 90 before beta decay.

Therefore, the parent nucleus is \38}^{90}Sr\].

Hence, the decay equation will be:

\[_{38}^{90}S{r_{52}} \to _{39}^{90}{Y_{51}}{ + _{ - 1}}\beta + {\bar v_e}\]

Therefore, the equation is: \[_{38}^{90}S{r_{52}} \to _{39}^{90}{Y_{51}}{ + _{ - 1}}\beta + {\bar v_e}\].