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Chapter 25: Problem 1

What nucleus is obtained in each process? (a) \(\frac{234}{94}\) Pu decays by \(\alpha\) emission. \(\text { (b) } \begin{array}{l}248 \\\97\end{array}\)Bk decays by \(\beta^{-}\) emission. \(\text { (c) } \begin{array}{r}196 \\\82\end{array}\) Pb goes through two successive EC processes.

Short Answer

Expert verified
(a) The final nucleus after alpha decay of Pu is Thorium (Th), (b) The final nucleus after beta decay of Bk is Californium (Cf), (c) The final nucleus after two successive electron captures of Pb is Mercury (Hg).

Step by step solution

01

Alpha Decay of Pu

In alpha decay, the atomic number decreases by 2 and the mass number decreases by 4. So, \( \frac{234}{94}\) Pu decays by alpha emission to \( \frac{230}{92}\). Therefore, the resultant nucleus is Thorium (Th). This finding is based on the Periodic Table which shows that the element with an atomic number 92 is Thorium (Th).
02

Beta Decay of Bk

In beta decay, a neutron in the atomic nucleus is converted into a proton which increases the atomic number by 1 (mass number stays the same). So, \( \frac{248}{97}\) Bk decays by beta emission to \( \frac{248}{98}\). Therefore, the resultant nucleus is Californium (Cf), as per the Periodic Table. The element with an atomic number 98 is Californium (Cf).
03

Electron Capture of Pb

In electron capture (EC), a proton in the nucleus is converted into a neutron which decreases the atomic number by 1 (mass number stays the same). Since Pb goes through two successive EC processes, the atomic number will decrease by 2 in total. So \( \frac{196}{82}\) Pb after two successive EC reactions becomes \( \frac{196}{80}\). Hence, the resultant nucleus is Mercury (Hg), following the Periodic Table. The element with an atomic number 80 is Mercury (Hg).

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Most popular questions from this chapter

Write nuclear equations to represent the formation of a hypothetical isotope of element 118 with a mass number of 293 by the bombardment of lead- 208 by krypton-86 nuclei, followed by a chain of \(\alpha\) -particle emissions to the element seaborgium.

Write nuclear equations to represent the formation of an isotope of element 111 with a mass number of 272 by the bombardment of bismuth-209 by nickel-64 nuclei, followed by a succession of five \(\alpha\) -particle emissions.

For medical uses, radon-222 formed in the radioactive decay of radium-226 is allowed to collect over the radium metal. Then, the gas is withdrawn and sealed into a glass vial. Following this, the radium is allowed to disintegrate for another period, when a new sample of radon- 222 can be withdrawn. The procedure can be continued indefinitely. The process is somewhat complicated by the fact that radon-222 itself undergoes radioactive decay to polonium- 218 , and so on. The half-lives of radium-226 and radon-222 are \(1.60 \times 10^{3}\) years and 3.82 days, respectively.(a) Beginning with pure radium- \(226,\) the number of radon-222 atoms present starts at zero, increases for a time, and then falls off again. Explain this behavior. That is, because the half-life of radon-222 is so much shorter than that of radium- \(226,\) why doesn't the radon-222 simply decay as fast as it is produced, without ever building up to a maximum concentration?(b) Write an expression for the rate of change \((d \mathrm{D} / d t)\) in the number of atoms (D) of the radon- 222 daughter in terms of the number of radium- 226 atoms present initially ( \(\mathrm{P}_{0}\) ) and the decay constants of the parent \(\left(\lambda_{\mathrm{p}}\right)\) and daughter \(\left(\lambda_{\mathrm{d}}\right)\) (c) Integration of the expression obtained in part (b) yields the following expression for the number of atoms of the radon-222 daughter (D) present at a time \(t\).$$\mathrm{D}=\frac{\mathrm{P}_{0} \lambda_{\mathrm{p}}\left(\mathrm{e}^{-\lambda_{\mathrm{p}} \times t}-\mathrm{e}^{-\lambda_{\mathrm{d}} \times t}\right)}{\lambda_{\mathrm{d}}-\lambda_{\mathrm{p}}}$$,Starting with \(1.00 \mathrm{g}\) of pure radium- \(226,\) approximately how long will it take for the amount of radon222 to reach its maximum value: one day, one week, one year, one century, or one millennium?

A lunar rock was analyzed for argon by mass spectrometry and for potassium by atomic absorption. The results of these analyses showed that the sample contained \(3.02 \times 10^{-5} \mathrm{mL} \mathrm{g}^{-1}\) of argon and \(0.083 \%\) of potassium. The half-life of potassium- 40 is \(1.248 \times\) \(10^{9} \mathrm{y} \cdot\) Calculate the age of the lunar rock.

A process that produces a one-unit increase in atomic number is (a) electron capture; (b) \(\beta^{-}\) emission;(c) \(\alpha\) emission; (d) \(\gamma\) -ray emission.
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