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Chapter 21: Problem 23

Fill in the blanks in these radioactive decay series: (a) \(^{232} \mathrm{Th} \stackrel{\alpha}{\longrightarrow}\) _______ \(\stackrel{\beta}{\longrightarrow}\) ________ \(\stackrel{\beta}{\longrightarrow}{ }^{228} \mathrm{Th}\) (b) \({ }^{235} \mathrm{U} \stackrel{\alpha}{\longrightarrow}\) ________ \(\stackrel{\beta}{\longrightarrow}\) _________ \(\stackrel{\alpha}{\longrightarrow}^{227} \mathrm{Ac}\) (c) _______ \(\stackrel{\alpha}{\longrightarrow}{ }^{233} \mathrm{~Pa} \stackrel{\beta}{\longrightarrow}\) ___________ \(\stackrel{\alpha}{\longrightarrow}\) ________.

Short Answer

Expert verified
The filled decay series are: (a) \(^{232} \mathrm{Th} \stackrel{\alpha}{\longrightarrow}{ }^{228} \mathrm{Ra}\)\(\stackrel{\beta}{\longrightarrow}{ }^{228} \mathrm{Fr}\) \(\stackrel{\beta}{\longrightarrow}{ }^{228} \mathrm{Th}\), (b) \({ }^{235} \mathrm{U} \stackrel{\alpha}{\longrightarrow}{ }^{231} \mathrm{Th}\) \(\stackrel{\beta}{\longrightarrow}{ }^{231} \mathrm{Pa}\) \(\stackrel{\alpha}{\longrightarrow}^{227} \mathrm{Ac}\), (c) \({ }^{237} \mathrm{Th} \stackrel{\alpha}{\longrightarrow}{ }^{233} \mathrm{~Pa} \stackrel{\beta}{\longrightarrow}{ }^{233} \mathrm{Th}\) \(\stackrel{\alpha}{\longrightarrow}{ }^{229} \mathrm{Fr}\)

Step by step solution

01

Solve for (a)

In alpha decay, atomic number decreases by 2 and atomic mass decreases by 4. Thus, after the alpha decay of Thorium 232, we get Radon 228.After the beta decay, the atomic number increases by one and we get Francium 228.After the second beta decay, we again increase the atomic number by one so we finally get Thorium 228. Hence, the series is: \(^{232} \mathrm{Th} \stackrel{\alpha}{\longrightarrow}{ }^{228} \mathrm{Ra}\)\(\stackrel{\beta}{\longrightarrow}{ }^{228} \mathrm{Fr}\) \(\stackrel{\beta}{\longrightarrow}{ }^{228} \mathrm{Th}\)
02

Solve for (b)

Using the same logic we infer that after alpha decay from Uranium 235, we get Thorium 231.After the beta decay, the atomic number increases by one so, we get Protactinium 231.After another alpha decay, we get Actinium 227. Hence, the series is: \({ }^{235} \mathrm{U} \stackrel{\alpha}{\longrightarrow}{ }^{231} \mathrm{Th}\) \(\stackrel{\beta}{\longrightarrow}{ }^{231} \mathrm{Pa}\) \(\stackrel{\alpha}{\longrightarrow}^{227} \mathrm{Ac}\)
03

Solve for (c)

The series starts with an alpha decay. So we increase the atomic number by 2 and the mass by 4 to the Pa-233, obtaining Thorium 237.The Thorium 237 then undergoes a beta decay, transforming into Protactinium 237. Lastly, after another alpha decay, the result is Francium 233. Hence, the series is: \({ }^{237} \mathrm{Th} \stackrel{\alpha}{\longrightarrow}{ }^{233} \mathrm{~Pa} \stackrel{\beta}{\longrightarrow}{ }^{233} \mathrm{Th}\) \(\stackrel{\alpha}{\longrightarrow}{ }^{229} \mathrm{Fr}\).

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Most popular questions from this chapter

Tritium contains one proton and two neutrons. There is no proton-proton repulsion present in the nucleus. Why, then, is tritium radioactive?

Define nuclear fission, nuclear chain reaction, and critical mass.

(a) What is the activity, in millicuries, of a 0.500 -g sample of \({ }_{93}^{237} \mathrm{~Np}\) ? (This isotope decays by \(\alpha\) -particle emission and has a half-life of \(2.20 \times 10^{6}\) yr.) (b) Write a balanced nuclear equation for the decay of \({ }^{237} \mathrm{~Np}\)

How is nuclear transmutation achieved in practice?

Consider this redox reaction: $$ \begin{array}{r} \mathrm{IO}_{4}^{-}(a q)+2 \mathrm{I}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \\ \mathrm{I}_{2}(s)+\mathrm{IO}_{3}^{-}(a q)+2 \mathrm{OH}^{-}(a q) \end{array} $$ When \(\mathrm{KIO}_{4}\) is added to a solution containing iodide ions labeled with radioactive iodine- \(128,\) all the radioactivity appears in \(\mathrm{I}_{2}\) and none in the \(\mathrm{IO}_{3}^{-}\) ion. What can you deduce about the mechanism for the redox process?
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