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Chapter 21: Problem 72

Radon-222 decays to a stable nucleus by a series of three alpha emissions and two beta emissions. What is the stable nucleus that is formed?

Short Answer

Expert verified
The stable nucleus formed after the decay of Radon-222 consisting of three alpha emissions and two beta emissions is Lead-210 (\(^{210}_{82}Pb\)).

Step by step solution

01

Recall alpha and beta decay process

Before solving this exercise, it is necessary to recall the particle emissions in alpha and beta decays: 1. In alpha decay, an alpha particle (which is a Helium-4 nucleus, containing 2 protons and 2 neutrons) is emitted. This leads to the atomic number decreasing by 2 and the mass number decreasing by 4. 2. In beta decay, a beta particle (a high-energy electron) is emitted. This occurs when a neutron changes into a proton and an electron, causing the atomic number to increase by 1, while the mass number remains unchanged.
02

Write the initial Radon-222 isotope

Radon-222 has an atomic number of 86 and a mass number of 222. We represent the isotope as \(^{222}_{86}Rn\).
03

Perform the first alpha decay

In the first alpha decay, the atomic number decreases by 2 and the mass number decreases by 4. New isotope: \(^{218}_{84}Po\) (Polonium-218)
04

Perform the second alpha decay

In the second alpha decay, the atomic number decreases by 2 again and the mass number decreases by 4. New isotope: \(^{214}_{82}Pb\) (Lead-214)
05

Perform the third alpha decay

In the third alpha decay, the atomic number decreases by 2 and the mass number decreases by 4. New isotope: \(^{210}_{80}Hg\) (Mercury-210)
06

Perform the first beta decay

In the first beta decay, the atomic number increases by 1, and the mass number remains unchanged. New isotope: \(^{210}_{81}Tl\) (Thallium-210)
07

Perform the second beta decay

In the second beta decay, the atomic number increases by 1, and the mass number remains the same as in the previous step. New isotope: \(^{210}_{82}Pb\) (Lead-210) So, the stable nucleus formed after the decay of Radon-222 consisting of three alpha emissions and two beta emissions is Lead-210.

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Most popular questions from this chapter

Which of the following statements best explains why alpha emission is relatively common, but proton emission is extremely rare? \begin{equation}\begin{array}{l}{\text { (a) Alpha particles are very stable because of magic numbers }} \\ \quad {\text { of protons and neutrons. }} \\\ {\text { (b) Alpha particles occur in the nucleus. }} \\ {\text { (c) Alpha particles are the nuclei of an inert gas. }} \\ {\text { (d) An alpha particle has a higher charge than a proton. }}\end{array}\end{equation}

A portion of the Sun's energy comes from the reaction \begin{equation}4_{1}^{1} \mathrm{H} \longrightarrow_{2}^{4} \mathrm{He}+2_{1}^{0} \mathrm{e} \end{equation}which requires a temperature of \(10^{6}\) to \(10^{7} \mathrm{K}\) . Use the mass of the helium-4 nucleus given in Table 21.7 to determine how much energy is released per mol of hydrogen atoms.

Each of the following transmutations produces a radionuclide used in positron emission tomography (PET). (a) Inequations (i) and (ii), identify the species signified as "X." (b) In equation (iii), one of the species is indicated as "d." What do you think it represents? \begin{equation}\begin{array}{l}{(\mathrm{i})^{14} \mathrm{N}(\mathrm{p}, \alpha) \mathrm{X}} \\ {(\mathrm{ii})^{18} \mathrm{O}(\mathrm{p}, \mathrm{X})^{18} \mathrm{F}} \\ {\text { (iii) }^{14} \mathrm{N}(\mathrm{d}, \mathrm{n})^{15} \mathrm{O}}\end{array}\end{equation}

Complete and balance the nuclear equations for the following fission reactions: \begin{equation}(a) ^{2235} \mathrm{U}+_{0}^{1} \mathrm{n} \longrightarrow _{62}^{160} \mathrm{Sm}+ _{30}^{72} \mathrm{Zn}+_{0}^{1} \mathrm{n}\end{equation}\begin{equation} (b)^{239} \mathrm{Pu}+_{0}^{1} \mathrm{n} \longrightarrow _{58}^{144} \mathrm{Ce}+ 2 _{0}^{1} \mathrm{n} \end{equation}

The accompanying graph illustrates the decay of \(_{42}^{88} \mathrm{Mo}\) which decays via positron emission. (a) What is the half-life of the decay? (b) What is the rate constant for the decay?(c) What fraction of the original sample of \(_{42}^{88} \mathrm{Mo}\) remains after 12 \(\mathrm{min}\) ? (d) What is the product of the decay process? [Section 21.4\(]\)
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