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Chapter 21: Problem 17

The naturally occurring radioactive decay series that begins with \(_{92}^{235} \mathrm{U}\) stops with formation of the stable \(_{82}^{207} \mathrm{Pb}\) nucleus. The decays proceed through a series of alpha-particle and beta- particle emissions. How many of each type of emission are involved in this series?

Short Answer

Expert verified
In the radioactive decay series starting with \(_{92}^{235}\mathrm{U}\) and ending with \(_{82}^{207}\mathrm{Pb}\), there are 7 alpha-particle emissions (A) and 4 beta-particle emissions (B).

Step by step solution

01

Identify the initial and final nuclei and their atomic and mass numbers.

The initial nucleus is \(_{92}^{235}\mathrm{U}\), with an atomic number of 92 and a mass number of 235. The final nucleus is \(_{82}^{207}\mathrm{Pb}\), with an atomic number of 82 and a mass number of 207.
02

Understand the changes in atomic and mass numbers during alpha-particle and beta-particle emissions.

In an alpha-particle emission, the atomic number of the nucleus decreases by 2, and the mass number decreases by 4. In a beta-particle emission, the atomic number increases by 1, while the mass number remains unchanged.
03

Determine the change in atomic and mass numbers in the decay series.

The change in atomic number is the final atomic number minus the initial atomic number, so it is 82 - 92 = -10. The change in mass number is the final mass number minus the initial mass number, so it is 207 - 235 = -28.
04

Set up equations for the number of alpha-particle and beta-particle emissions.

Let A be the number of alpha-particle emissions, and B be the number of beta-particle emissions. From Step 2, we know that alpha-particle emissions cause a decrease of 2 in the atomic number and 4 in the mass number, while beta-particle emissions cause an increase of 1 in the atomic number. Therefore, we can set up the following equations: -2A + B = -10 (change in atomic number) -4A = -28 (change in mass number)
05

Solve the two equations for A and B.

From the second equation, we can directly find the number of alpha-particle emissions: -4A = -28 A = 7 Now we can substitute the value of A in the first equation: -2(7) + B = -10 -14 + B = -10 B = 4
06

Interpret the results.

In the radioactive decay series starting with \(_{92}^{235}\mathrm{U}\) and ending with \(_{82}^{207}\mathrm{Pb}\), there are 7 alpha-particle emissions and 4 beta-particle emissions.

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Most popular questions from this chapter

Which type or types of nuclear reactors have these characteristics? \(\begin{array}{l}{\text { (a) Can use natural uranium as a fuel }} \\ {\text { (b) Does not use a moderator }} \\ {\text { (c) Can be refueled without shutting down }}\end{array}\)

Complete and balance the following nuclear equations by supplying the missing particle: \begin{equation}\begin{array}{l}{\text { (a) }_{98}^{252} \mathrm{Cf}+_{5}^{10} \mathrm{B} \longrightarrow 3_{0}^{1} \mathrm{n}+?} \\\ {\text { (b) }_{1}^{2} \mathrm{H}+_{2}^{3} \mathrm{He} \longrightarrow_{2}^{4} \mathrm{He}+?}\\\ {\text { (c) }_{1}^{1} \mathrm{H}+_{5}^{11} \mathrm{B} \longrightarrow 3?} \\ {\text { (d) }_{53}^{122} \mathrm{I}\longrightarrow_{54}^{122} \mathrm{Xe}+?}\\\ {\text { (e) }_{26}^{59} \mathrm{Fe}\longrightarrow_{-1}^{0} \mathrm{e}+?}\end{array}\end{equation}

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?

Tests on human subjects in Boston in 1965 and 1966, following the era of atomic bomb testing, revealed average quantities of about 2 pCi of plutonium radioactivity in the average person. How many disintegrations per second does this level of activity imply? If each alpha particle deposits \(8 \times 10^{-13} \mathrm{J}\) of energy and if the average person weighs 75 kg, calculate the number of rads and rems of radiation in 1 yr from such a level of plutonium.

Nuclear scientists have synthesized approximately 1600 nuclei not known in nature. More might be discovered with heavy-ion bombardment using high-energy particle accelerators. Complete and balance the following reactions, which involve heavy-ion bombardments: \begin{equation} (a) \stackrel{6}{3} \mathrm{Li}+\stackrel{56}{28} \mathrm{Ni} \longrightarrow ? \end{equation}\begin{equation}(b) \stackrel{40}{20} \mathrm{Ca}+\stackrel {248}{96} \mathrm{Cm} \longrightarrow \stackrel{147}{62} \mathrm{Sm} + ? \end{equation}\begin{equation}(c) \stackrel{88}{38} \mathrm{Sr}+\stackrel{84}{36} \mathrm{Kr} \longrightarrow \stackrel{116}{46} \mathrm{Pd} + ?\end{equation}\begin{equation} (d)\stackrel{40}{20} \mathrm{Ca}+\stackrel{238}{92} \mathrm{U} \longrightarrow \stackrel{70}{30} \mathrm{Zn}+4 \stackrel{1}{0}\mathrm{n}+2 ?\end{equation}
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