Write a nuclear \(\beta^{-}\) decay reaction that produces the \(^{90} \mathrm{Y}\) nucleus. (Hint: The parent nuclide is a major waste product of reactors and has chemistry similar to calcium, so that it is concentrated in bones if ingested.)

Based on the hint provided, we know that the parent nuclide has similar chemistry to calcium and is a major waste product of reactors. We can identify this nuclide as \(^{90}\mathrm{Sr}\), which is a well-known radioisotope produced in nuclear reactors.

In β⁻ decay, a neutron in the nucleus is converted into a proton, an electron (also known as a beta particle), and an electron antineutrino. The emitted beta particle and the electron antineutrino carry away some energy from the nucleus. We can represent this conversion in the neutron-to-proton transformation process using the following reaction: \[n \rightarrow p + e⁻ + \bar{\nu}_e\]

Now that we know the parent nuclide \(^{90}\mathrm{Sr}\) and the fundamental process of β⁻ decay, we can write the nuclear β⁻ decay equation. As \(^{90}\mathrm{Sr}\) undergoes β⁻ decay, one of its neutrons will convert into a proton, increasing the atomic number (which identifies the element) by 1 while the mass number remains the same. This will result in the formation of the \(^{90}\mathrm{Y}\) nucleus. The full decay equation is as follows: \[^{90}\mathrm{Sr} \rightarrow ^{90}\mathrm{Y} + e⁻ + \bar{\nu}_e\] In conclusion, we have written the nuclear β⁻ decay reaction that produces the \(^{90}\mathrm{Y}\) nucleus starting from the parent nuclide \(^{90}\mathrm{Sr}\). The final equation representing this reaction is given by: \[^{90}\mathrm{Sr} \rightarrow ^{90}\mathrm{Y} + e⁻ + \bar{\nu}_e\]