What particle is produced during the following decay processes: \((\mathbf{a})\) sodium-24 decays to magnesium-24; \((\mathbf{b})\) mercury-188 decays to gold-188; \((\mathbf{c})\)iodine-122 decays to xenon-122; \((\mathbf{d})\) plutonium-242 decays to uranium-238?

Short Answer

Expert verified
In the given decay processes: \((\mathbf{a})\) sodium-24 to magnesium-24 and \((\mathbf{b})\) mercury-188 to gold-188, both produce an electron \(\beta^{-}\) (a beta-minus particle) since they are beta decays. \((\mathbf{c})\) iodine-122 to xenon-122 produces a positron \(\beta^{+}\) (a beta-plus particle) due to positron emission. And finally, \((\mathbf{d})\) plutonium-242 to uranium-238 produces an alpha particle (_2^4He) as it is an alpha decay process.

Step by step solution

01

Identify the decay process in each case

In order to identify the particles produced, we need to first identify the decay process occurring in each case. \((\mathbf{a})\) Sodium-24 decays to magnesium-24: This is a beta decay process, as the atomic number (number of protons) increases by 1 while the atomic mass (number of protons and neutrons) remains the same. \((\mathbf{b})\) Mercury-188 decays to gold-188: This is also a beta decay process, as the atomic number increases by 1, and the atomic mass remains the same. \((\mathbf{c})\) Iodine-122 decays to xenon-122: In this decay process, the atomic number decreases by 1 while the atomic mass remains the same. This is a positron emission process (beta-plus decay). \((\mathbf{d})\) Plutonium-242 decays to uranium-238: This decay process involves the atomic number decreasing by 2, and the atomic mass decreasing by 4. This indicates an alpha decay process. Now that the decay processes are identified, we can determine the particles produced during each decay.
02

Identify the particle produced in each decay process

Based on the type of decay identified in Step 1, we can now identify the particle produced in each decay process. \((\mathbf{a})\) Sodium-24 decays to magnesium-24: Beta decay produces an electron \(\beta^{-}\) (a beta-minus particle). \((\mathbf{b})\) Mercury-188 decays to gold-188: Beta decay produces an electron \(\beta^{-}\) (a beta-minus particle). \((\mathbf{c})\) Iodine-122 decays to xenon-122: Positron emission produces a positron \(\beta^{+}\) (a beta-plus particle). \((\mathbf{d})\) Plutonium-242 decays to uranium-238: Alpha decay produces an alpha particle, which consists of 2 protons and 2 neutrons, represented by \(_2^4\text{He}\). And there we have it! We've identified the particle produced during each decay process: an electron (\(\beta^{-}\)) for \((\mathbf{a})\) and \((\mathbf{b})\), a positron (\(\beta^{+}\)) for \((\mathbf{c})\), and an alpha particle for \((\mathbf{d})\).

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Most popular questions from this chapter

Chlorine has two stable nuclides, \(^{35} \mathrm{Cl}\) and \(^{37} \mathrm{Cl} .\) In contrast, \(^{36} \mathrm{Cl}\) is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of \(^{36} \mathrm{Cl} ?\) (b) Based on the empirical rules about nuclear stability, explain why the nucleus of \(^{36} \mathrm{C}\) is less stable than either \(^{35}\mathrm{Cl}\) or \(^{37} \mathrm{Cl}\).

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