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Chapter 21: Problem 14

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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