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Chapter 21: Problem 25

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

Short Answer

Expert verified
In the radioactive decay series starting with \({ }_{90}^{232}\) Th and ending with \({ }_{82}^{208}\mathrm{Pb}\), there are 6 alpha-particle emissions and 20 beta-particle emissions.

Step by step solution

01

Calculate the change in atomic number and atomic mass

To find how many alpha and beta emissions occurred, we need to calculate the difference in atomic number and atomic mass between the initial and final nuclides. Initial nuclide: \({ }_{90}^{232}\) Th Final nuclide: \({ }_{82}^{208}\mathrm{Pb}\) Change in atomic number: \(\Delta Z = 90 - 82 = 8\) Change in atomic mass: \(\Delta A = 232 - 208 = 24\)
02

Find the number of alpha decays

Since each alpha decay reduces the atomic mass by 4, we can calculate the number of alpha decays by dividing the total change in atomic mass by 4: Number of alpha decays = \(\frac{\Delta A}{4} = \frac{24}{4} = 6\)
03

Find the number of beta decays

Since each alpha decay reduces the atomic number by 2 and each beta decay increases the atomic number by 1, we can calculate the number of beta decays by using the fact that the total number of alpha and beta decays results in the observed change in atomic number. Let \(a\) be the number of alpha decays and \(b\) be the number of beta decays, then: \(-2a + b = \Delta Z\) We already found the number of alpha decays (\(a = 6\)), so we can plug that back into the equation: \(-2(6) + b = 8\) Solving for \(b\): \(b = 8 + 2(6) = 20\)
04

Conclusion

In the radioactive decay series starting with \({ }_{90}^{232}\) Th and ending with \({ }_{82}^{208}\mathrm{Pb}\), there are 6 alpha-particle emissions and 20 beta-particle emissions.

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Most popular questions from this chapter

Chlorine has two stable nuclides, \({ }^{35} \mathrm{Cl}\) and \({ }^{37} \mathrm{Cl}\). In contrast, \({ }^{36} \mathrm{Cl}\) is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of \({ }^{36} \mathrm{Cl}\) ? (b) Based on the empirical rules about nuclear stability, explain why the nucleus of \({ }^{36} \mathrm{Cl}\) is less stable than either \({ }^{35} \mathrm{Cl}\) or \({ }^{37} \mathrm{Cl}\).

A wooden artifact from a Chinese temple has a \({ }^{14} \mathrm{C}\) activity of \(38.0\) counts per minute as compared with an activity of \(58.2\) counts per minute for a standard of zero age. From the halflife for \({ }^{14} \mathrm{C}\) decay, \(5715 \mathrm{yr}\), determine the age of the artifact.

Naturally found uranium consists of \(99.274 \%^{238} \mathrm{U}\), \(0.720 \%{ }^{235} \mathrm{U}\), and \(0.006 \%{ }^{233} \mathrm{U}\). As we have \(\sec ,{ }^{235} \mathrm{U}\) is the isotope that can undergo a nuclear chain reaction. Most of the \({ }^{235} \mathrm{U}\) used in the first atomic bomb was obtained by gascous diffusion of uranium hexafluoride, \(\mathrm{UF}_{6}(g)\). (a) What is the mass of \(\mathrm{UF}_{6}\) in a \(30.0-\mathrm{L}\) vessel of \(\mathrm{UF}_{6}\) at a pressure of 695 torr at \(350 \mathrm{~K}\) ? (b) What is the mass of \({ }^{235} \mathrm{U}\) in the sample described in part (a)? (c) Now suppose that the \(\mathrm{UF}_{6}\) is diffused through a porous barricr and that the change in the ratio of \({ }^{23} \mathrm{U}\) and \({ }^{2 / 4} \mathrm{U}\) in the diffused gas can he deseribed by Equation 10.23. What

A rock contains \(0.257 \mathrm{mg}\) of lead-206 for every milligram of uranium-238. The half-life for the decay of uranium-238 to lead-206 is \(4.5 \times 10^{5} \mathrm{yr}\). How old is the rock? SOLUTION Analyze We are told that a rock sample has a certain amount of lead206 for every unit mass of uranium-238 and asked to estimate the age of the rock. Plan Lead-206 is the product of the radioactive decay of uranium-238. We will assume that the only source of lead-206 in the rock is from the decay of uranium-238, with a known half-life. To apply firstorder kinetics expressions (Equations \(21.19\) and 21.20) to calculate the time elapsed since the rock was formed, we first need to calculate how much initial uranium-238 there was for every \(1 \mathrm{mg}\) that remains today. Solve Let's assume that the rock currently contains \(1.000 \mathrm{mg}\) of uranium-238 and therefore \(0.257 \mathrm{mg}\) of lead-206. The amount of uranium-238 in the rock when it was first formed therefore equals \(1.000 \mathrm{mg}\) plus the quantity that has decayed to lead-206. Because the mass of lead atoms is not the same as the mass of uranium atoms, we cannot just add \(1.000 \mathrm{mg}\) and \(0.257 \mathrm{mg}\). We have to multiply the present mass of lead-206 \((0.257 \mathrm{mg})\) by the ratio of the mass number of uranium to that of lead, into which it has decayed. Therefore, the original mass of \({ }_{92}^{239} \mathrm{U}\) was $$ \text { Original } \begin{aligned} { }_{98}^{238} \mathrm{U} &=1.000 \mathrm{mg}+\frac{238}{206}(0.257 \mathrm{mg}) \\ &=1.297 \mathrm{mg} \end{aligned} $$ Using Equation 21.20, we can calculate the decay constant for the process from its half-life: $$ k=\frac{0.693}{4.5 \times 10^{9} \mathrm{yr}}=1.5 \times 10^{-10} \mathrm{yr}^{-1} $$ Rearranging Equation \(21.19\) to solve for time, \(t\), and substituting known quantities gives $$ t=-\frac{1}{k} \ln \frac{N_{t}}{N_{0}}=-\frac{1}{1.5 \times 10^{-10} \mathrm{yr}^{-1}} \ln \frac{1.000}{1.297}=1.7 \times 10^{9} \mathrm{yr} $$ Comment To check this result, you could use the fact that the decay of uranium-235 to lead-207 has a half-life of \(7 \times 10^{8} \mathrm{yr}\) and measure the relative amounts of uranium-235 and lead-207 in the rock.

Each statement that follows refers to a comparison between two radioisotopes, \(A\) and \(X\). Indicate whether each of the following statements is true or false, and why. (a) If the half-life for \(\mathrm{A}\) is shorter than the half-life for \(\mathrm{X}, \mathrm{A}\) has a larger decay rate constant. (b) If \(X\) is "not radioactive," its half-life is essentially zero. (c) If A has a half-life of \(10 \mathrm{yr}\), and \(\mathrm{X}\) has a half-life of \(10,000 \mathrm{yr}\), A would be a more suitable radioisotope to measure processes occurring on the 40 -yr time scale.
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