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Chapter 21: Problem 18

A radioactive decay series that begins with \(_{90}^{232}\) \(\mathrm{Th}\) ends with formation of the stable nuclide \(_{82}^{208}\) \(\mathrm{Pb}\) . How many alpha-particle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays?

Short Answer

Expert verified
In the radioactive decay series from \(_{90}^{232}\) Th to \(_{82}^{208}\) Pb, there are 4 alpha-particle emissions and no beta-particle emissions involved.

Step by step solution

01

Analyze the initial and final isotopes

We begin with the isotope \(_{90}^{232}\) Th and we end with the stable isotope \(_{82}^{208}\) Pb. The changes in atomic number (ΔZ) and mass number (ΔA) during the decay sequence can be expressed as: ΔZ = 90 - 82 = 8 ΔA = 232 - 208 = 24
02

Calculate the number of alpha emissions in the decay series

Each alpha decay reduces the mass number (A) by 4 and the atomic number (Z) by 2. Since the change in atomic number ΔZ is 8 and the change in mass number ΔA is 24, we can write an equation to calculate the number of alpha emissions (n_alpha) involved in the decay sequence: ΔZ = 2 * n_alpha 8 = 2 * n_alpha n_alpha = 4 Therefore, there are 4 alpha-particle emissions in the decay sequence.
03

Calculate the number of beta emissions in the decay series

Each beta decay increases the atomic number (Z) by 1 but does not change the mass number (A). We already determined that the change in atomic number ΔZ is 8. Since we have 4 alpha-particle emissions, which reduce the atomic number by 2 each, we need to calculate the number of beta emissions (n_beta) that compensate this change: ΔZ = 2 * n_alpha - n_beta 8 = 2 * 4 - n_beta 8 = 8 - n_beta n_beta = 0 Therefore, there are no beta-particle emissions involved in this decay sequence.
04

Summarize the results

In this radioactive decay series, there are 4 alpha-particle emissions and no beta-particle emissions involved in the decay sequence from \(_{90}^{232}\) Th to \(_{82}^{208}\) Pb.

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