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Chapter 21: Problem 24

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 decay series of Uranium-235 (\(^{235}_{92} \mathrm{U}\)) to Lead-207 (\(^{207}_{82} \mathrm{Pb}\)), there are 7 alpha decays and 4 beta decays.

Step by step solution

01

Understand Alpha and Beta Decay

During an alpha decay, an unstable nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons. This results in a decrease in the atomic number (Z) by 2 and a decrease in the mass number (A) by 4. During a beta decay, a neutron in the nucleus gets converted into a proton, and an electron (beta particle) is emitted. This results in an increase in the atomic number (Z) by 1, while the mass number (A) remains unchanged. Now that we understand how the atomic and mass numbers change during alpha and beta decays, we can proceed to determine the number of alpha and beta particle emissions in the given decay series.
02

Calculate the change in Atomic Number and Mass Number

The initial atomic number (Z1) and mass number (A1) of Uranium-235 are 92 and 235, respectively. The final atomic number (Z2) and mass number (A2) of Lead-207 are 82 and 207, respectively. The change in atomic number 螖Z = Z2 - Z1 = 82 - 92 = -10 The change in mass number 螖A = A2 - A1 = 207 - 235 = -28
03

Determine the number of Alpha and Beta emissions

Let 'a' be the number of alpha decays and 'b' be the number of beta decays. From Step 2, we know that after all the decays, the change in atomic number is -10 and change in mass number is -28. Since alpha decay decreases the atomic number by 2 and beta decay increases the atomic number by 1, we can write the equation: -2a + b = -10 Alpha decay also decreases the mass number by 4, while beta decay doesn't change the mass number. Hence, the equation becomes: -4a = -28 Now, we can solve the system of equations to find the values of 'a' and 'b'.
04

Solve the system of equations

From the second equation, we can find the number of alpha decays: -4a = -28 a = 7 Substitute the value of 'a' in the first equation: -2(7) + b = -10 -14 + b = -10 b = 4 Therefore, there are 7 alpha decays (a = 7) and 4 beta decays (b = 4) in the decay series of Uranium-235 to Lead-207.

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Most popular questions from this chapter

Write balanced nuclear equations for the following processes: (a) rubidium-90 undergoes beta emission; (b) selenium-72 undergoes electron capture; (c) krypton- 76 undergoes positron emission; (d) radium-226 emits alpha radiation.

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Iodine-131 is a convenient radioisotope to monitor thyroid activity in humans. It is a beta emitter with a half-life of \(8.02\) days. The thyroid is the only gland in the body that uses iodine. A person undergoing a test of thyroid activity drinks a solution of Nal, in which only a small fraction of the iodide is radioactive. (a) Why is NaI a good choice for the source of iodine? (b) If a Geiger counter is placed near the person's thyroid (which is near the neck) right after the sodium iodide solution is taken, what will the data look like as a function of time? (c) A normal thyroid will take up about \(12 \%\) of the ingested iodide in a few hours. How long will it take for the radioactive iodide taken up and held by the thyroid to decay to \(0.01 \%\) of the original amount?

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
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