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Chapter 29: Problem 24

Write out the reaction and identify the daughter nuclide when \(_{11}^{22}\) Na decays by emitting a positron.

Short Answer

Expert verified
Answer: The daughter nuclide is \(_{10}^{22}\)Ne, and the reaction is given by the balanced nuclear equation: \(_{11}^{22}\)Na \(\rightarrow\) \(_{10}^{22}\)Ne + \(_1^0\)e + ν.

Step by step solution

01

Understand positron emission

Positron emission, or beta-plus decay, is a type of radioactive decay in which a proton in the nucleus is converted into a neutron, releasing a positron and a neutrino. A positron is an anti-electron, which means it has the same mass as an electron but a positive charge. In terms of the nuclide, this process causes the atomic number to decrease by 1, while the mass number remains unchanged.
02

Write the initial nuclear equation

To determine the daughter nuclide, we will first write down the initial nuclear equation, where the left side represents the parent nuclide and the right side shows what is produced in the decay process. The parent nuclide is given as \(_{11}^{22}\)Na, and since it is undergoing positron emission, we write a positron, indicated by \(_1^0\)e (or β⁺), and a neutrino, indicated by ν. Initial nuclear equation: \(_{11}^{22}\)Na \(\rightarrow\) Daughter nuclide + \(_1^0\)e + ν
03

Determine the daughter nuclide

As mentioned before, in positron emission, the atomic number decreases by 1 while the mass number remains the same. Therefore, the atomic number of the daughter nuclide will be 11-1 = 10, and the mass number will be 22. The element with an atomic number of 10 is Neon (Ne), so the daughter nuclide is \(_{10}^{22}\)Ne. Now we can rewrite the balanced nuclear equation.
04

Write the balanced nuclear equation

Now that we have identified the daughter nuclide, we can write the balanced nuclear equation: Balanced nuclear equation: \(_{11}^{22}\)Na \(\rightarrow\) \(_{10}^{22}\)Ne + \(_1^0\)e + ν The correct answer is that when \(_{11}^{22}\)Na decays by emitting a positron, \(_{10}^{22}\)Ne is the daughter nuclide, and the reaction is given by the balanced nuclear equation above.

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Most popular questions from this chapter

\(_{52}^{106} \mathrm{Te}\) is radioactive; it \(\alpha\) decays to $^{102} \mathrm{s}_{0} \mathrm{Sn} .\( " \)_{50}^{102} \mathrm{Sn}$ is itself radioactive and has a half-life of 4.5 s. At \(t=0,\) a sample contains 4.00 mol of \(^{106}\) Te and 1.50 mol of \(^{102}\) so Sn. At \(t=\) \(25 \mu \mathrm{s},\) the sample contains \(3.00 \mathrm{mol}\) of \(^{106} \mathrm{Te}\) and $2.50 \mathrm{mol}\( of \)^{102} \mathrm{s} 0$ Sn. How much 102 50 Sn will there be at \(t=50 \mu \mathrm{s} ?\)
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