Question

Chromium- 51 , a positron emitter, is used in research to study red cell survival. It is delivered as a solution of sodium chromate. Write the nuclear equation for the decay of \(\mathrm{Cr}-51\).

Short answer

The nuclear equation for the decay of Cr-51 through positron emission is: $$_{24}^{51}\mathrm{Cr} \rightarrow _{23}^{51}\mathrm{V} + _{+1}^{0}\mathrm{e} \text{ or } _{1}^{0}\beta^{+}$$ This equation represents the transformation of Chromium-51 into Vanadium-51 and the emission of a positron particle.

Step-by-step solution

Identify the decay process and initial reactant

Since we are given that Chromium-51 undergoes positron emission (also known as beta plus decay), the initial reactant for this nuclear equation is Cr-51, which can be represented as \(_{24}^{51}\mathrm{Cr}\).

Identify the emitted particle

In positron emission, an unstable nucleus emits a positron particle to reduce the number of protons and increase the number of neutrons. A positron has the same mass as an electron, but with a positive charge. It can be represented as \(_{+1}^{0}\mathrm{e}\) or \(_{1}^{0}\beta^{+}\).

Determine the resulting product

After the emission of a positron, the atomic number of the radionuclide decreases by one, while the mass number remains the same. In this case, the atomic number decreases from 24 to 23, but the mass number stays at 51. Using the periodic table, we can see that the element with an atomic number of 23 is Vanadium (V). So, the resulting product is \(_{23}^{51}\mathrm{V}\).

Write the nuclear equation for Cr-51 decay

Now that we have identified the initial reactant, emitted particle, and resulting product, we can write the nuclear equation for the decay of Cr-51: $$_{24}^{51}\mathrm{Cr} \rightarrow _{23}^{51}\mathrm{V} + _{+1}^{0}\mathrm{e} \text{ or } _{1}^{0}\beta^{+}$$ This equation represents the nuclear decay of Cr-51 through positron emission, resulting in the formation of V-51 and a positron.