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Chapter 21: Problem 15

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
There are 7 alpha decays and 4 beta decays involved in the radioactive decay series from \(\mathrm{^{235}U}\) to \(\mathrm{^{207}Pb}\).

Step by step solution

01

List down the initial and final isotopes

The initial isotope is \(\mathrm{^{235}U}\) with 92 protons and 235 nucleons. The final stable isotope is \(\mathrm{^{207}Pb}\) with 82 protons and 207 nucleons.
02

Write down the alpha and beta decay process

In an alpha decay, a nucleus loses 2 protons and 2 neutrons, represented by an alpha particle, which is an \(\mathrm{^4He}\) nucleus: $$_{Z}^{A}X \rightarrow _{Z-2}^{A-4}Y + _2^4\mathrm{He}$ In a beta decay, a neutron in the nucleus is converted into a proton, while emitting a beta particle, which is an electron: $$_{Z}^{A}X \rightarrow _{Z+1}^{A}Y + _{-1}^0\mathrm{e}$
03

Set up a system of linear equations

Let x be the number of alpha decays and y be the number of beta decays. After x alpha decays, the number of protons decreases by 2x and the number of nucleons decreases by 4x. After y beta decays, the number of protons increases by y and the number of nucleons remains the same. Now, we can set up a system of equations: 1. Protons: \(92 - 2x + y = 82\) 2. Nucleons: \(235 - 4x = 207\)
04

Solve the system of linear equations

First, let's solve Equation 2 for x: \(4x = 235 - 207\) \(x = (235-207)/4 = 7\) Now, substitute x in Equation 1: \(92 - 2(7) + y = 82\) \(92 - 14 + y = 82\) \(y = 82 - (92-14) = 4\) So, there are 7 alpha decays and 4 beta decays involved in this decay series.

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Most popular questions from this chapter

The thermite reaction, \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Fe}(s)+\) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s), \Delta H^{\circ}=-851.5 \mathrm{~kJ} / \mathrm{mol},\) is one of the most exother-mic reactions known. Because the heat released is sufficient to melt the iron product, the reaction is used to weld metal under the ocean. How much heat is released per mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) produced? How does this amount of thermal energy compare with the energy released when 2 mol of protons and 2 mol of neutrons combine to form 1 mol of alpha particles?

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