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What particle is produced during the following decay processes: (a) sodium- 24 decays to magnesium- \(24 ;\) (b) mercury188 decays to gold-188; (c) iodine-122 decays to xenon-122; (d) plutonium-242 decays to uranium-238?

Short Answer

Expert verified
In the given decay processes: (a) a positron (\( _{1}^{0}e^{+} \)) is produced when sodium-24 decays to magnesium-24; (b) an electron (\( _{-1}^{0}e^{-} \)) is produced when mercury-188 decays to gold-188; (c) a positron (\( _{1}^{0}e^{+} \)) is produced when iodine-122 decays to xenon-122; and (d) an alpha particle (\( _{-2}^{-4}\alpha \)) is produced when plutonium-242 decays to uranium-238.

Step by step solution

01

(a) Sodium-24 decays to Magnesium-24

To solve this, we first need to write down the isotopes in their respective nuclear notation: \( _{11}^{24}Na \rightarrow _{12}^{24}Mg + X \) Now, we'll compare the atomic and mass numbers to find X: Atomic numbers: 12 - 11 = 1 Mass numbers: 24 - 24 = 0 Therefore, X has an atomic number of 1 and a mass number of 0, which corresponds to a positron (anti-electron): \(\rightarrow _{12}^{24}Mg + _{1}^{0}e^{+} \)
02

(b) Mercury-188 decays to Gold-188

We will follow the same approach as in part (a): \( _{80}^{188}Hg \rightarrow _{79}^{188}Au + X \) Comparing the atomic and mass numbers to find X: Atomic numbers: 79 - 80 = -1 Mass numbers: 188 - 188 = 0 Therefore, X has an atomic number of -1 and a mass number of 0, which corresponds to an electron: \(\rightarrow _{79}^{188}Au + _{-1}^{0}e^{-} \)
03

(c) Iodine-122 decays to Xenon-122

Following the same approach: \( _{53}^{122}I \rightarrow _{54}^{122}Xe + X \) Comparing the atomic and mass numbers to find X: Atomic numbers: 54 - 53 = 1 Mass numbers: 122 - 122 = 0 Therefore, X has an atomic number of 1 and a mass number of 0, which corresponds to a positron (anti-electron): \(\rightarrow _{54}^{122}Xe + _{1}^{0}e^{+} \)
04

(d) Plutonium-242 decays to Uranium-238

Following the same approach: \( _{94}^{242}Pu \rightarrow _{92}^{238}U + X \) Comparing the atomic and mass numbers to find X: Atomic numbers: 92 - 94 = -2 Mass numbers: 238 - 242 = -4 Therefore, X has an atomic number of -2 and a mass number of -4, which corresponds to an alpha particle (helium nucleus): \(\rightarrow _{92}^{238}U + _{-2}^{-4}\alpha \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Positron Emission
Positron emission is a type of radioactive decay in which a proton inside a nucleus is converted into a neutron, and a positron is released. In layman terms, a positron is basically the antimatter counterpart of an electron. It has the same mass as an electron but carries a positive charge. When isotopes like Sodium-24 decay into Magnesium-24, positron emission plays a crucial part in this nuclear transformation. You can identify positron emission by observing a change in atomic number without a change in mass number: the atomic number decreases by one because a proton becomes a neutron. This helps stabilize the nucleus by reducing the number of protons relative to neutrons. In nuclear notation, a positron emitted can be represented as \(_{1}^{0}e^{+}\), highlighting that it has an atomic number of 1 (indicative of a 'positive electron') but no mass number.
Electron Capture
Electron capture is another fascinating process within radioactive decay. Here, an unstable atom seizes an electron from its inner orbit, usually the K-shell, and incorporates it into the nucleus. This captured electron merges with a proton to form a neutron and emits a neutrino. Unlike positron emission, where a positron is ejected, electron capture doesn't release a particle that can be readily detected. This process effectively reduces the atomic number of the nucleus by one, as a proton within the nucleus is transformed into a neutron. However, like positron emission, the mass number remains unchanged. Electron capture can often be identified when analyzing decays where the atomic number decreases by one without any other particles emitted, as sometimes seen in writing nuclear equations.
Nuclear Notation
Nuclear notation is a concise way to represent isotopes and the changes they undergo during radioactive decay. Each element is represented by its atomic number (number of protons) and mass number (total number of protons and neutrons). For instance, Sodium-24 is written as \(_{11}^{24}Na\), making it easy to track what happens to the atomic and mass numbers during decay processes. Understanding nuclear notation is essential as it allows one to track and predict
  • the type of decay involved,
  • the resultant elements, and
  • any particles emitted.
By comparing changes in atomic number and mass, students can identify whether decay involves positron emission, electron capture, or alpha decay.
Alpha Decay
Alpha decay is a type of radioactive decay where an atomic nucleus ejects an alpha particle, consisting of two protons and two neutrons—essentially a helium nucleus. This significantly decreases the mass of the original element by 4 units and the atomic number by 2, as shown in the equation: \( _{94}^{242}Pu \rightarrow _{92}^{238}U + _{2}^{4}He \).Alpha particles are relatively heavy and positively charged, making them less penetrating than other forms of radiation but still quite energetic. Alpha decay typically occurs in heavy elements that don't have enough energy to become stable through other decay processes. It's considered one of the more predictable types of decay processes, easily recognized in nuclear notation by the reduction of two units in atomic number and four units in mass number—evidence of an emitted alpha particle.

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Most popular questions from this chapter

Predict the type of radioactive decay process for the following radionuclides: (a) \({ }_{5}^{8} \mathrm{~B},\) (b) \({ }_{29}^{68} \mathrm{Cu},\) (c) phosphorus-32, (d) chlorine- 39 .

Which of the following nuclides have magic numbers of both protons and neutrons: (a) helium- \(4,(\mathbf{b})\) oxygen \(-18,(\mathbf{c})\) calcium\(40,(\mathbf{d}) \operatorname{zinc}-66,(\mathbf{e})\) lead \(-208 ?\)

Hydroxyl radicals can pluck hydrogen atoms from molecules ("hydrogen abstraction"), and hydroxide ions can pluck protons from molecules ("deprotonation"). Write the reaction equations and Lewis dot structures for the hydrogen abstraction and deprotonation reactions for the generic carboxylic acid \(\mathrm{R}-\mathrm{COOH}\) with hydroxyl radical and hydroxide ion, respectively. Why is hydroxyl radical more toxic to living systems than hydroxide ion?

The thermite reaction, \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Fe}(s)+\) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s), \Delta H^{\circ}=-851.5 \mathrm{~kJ} / \mathrm{mol},\) is one of the most exother-mic reactions known. Because the heat released is sufficient to melt the iron product, the reaction is used to weld metal under the ocean. How much heat is released per mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) produced? How does this amount of thermal energy compare with the energy released when 2 mol of protons and 2 mol of neutrons combine to form 1 mol of alpha particles?

Write balanced equations for each of the following nuclear reactions: (a) \({ }_{92}^{238} \mathrm{U}(\mathrm{n}, \gamma){ }^{239} \mathrm{U}\) (b) \({ }_{7}^{14} \mathrm{~N}(\mathrm{p}, \alpha)^{11}{ }_{6} \mathrm{C}\) (c) \({ }_{8}^{18} \mathrm{O}(\mathrm{n}, \beta){ }^{19}{ }_{9} \mathrm{~F}\).

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