The steps below show three of the steps in the radioactive decay chain for \(_{90}^{232}\) Th. The half-life of each isotope is shown below the symbol of the isotope. (a) Identify the type of radioactive decay for each of the steps (i), (ii), and (iii). (b) Which of the isotopes shown has the highest activity? (c) Which of the isotopes shown has the lowest activity? (d) The next step in the decay chain is an alpha emission. What is the next isotope in the chain? [Sections 21.2 and 21.4]

Short Answer

Expert verified
(a) The types of radioactive decay are: (i) Alpha decay (ii) Beta decay (iii) Gamma decay (b) The isotope with the highest activity is Ac, with a half-life of 6.25 hours. (c) The isotope with the lowest activity is Th, with a half-life of 14.0 billion years. (d) The next isotope in the decay chain after an alpha emission is \(_{87}^{224}\text{Fr}\), francium.

Step by step solution

01

Identify radioactive decay types for each step

For each step in the decay chain, examine the change in atomic number and mass number to determine the type of decay. There are three types of radioactive decay we need to consider: alpha decay, beta decay, and gamma decay. (i) \(_{90}^{232}\text{Th} \rightarrow _{88}^{228}\text{Ra}\) In this step, the atomic number decreases by 2, and the mass number decreases by 4. This indicates an alpha decay. (ii) \(_{88}^{228}\text{Ra} \rightarrow _{89}^{228}\text{Ac}\) In this step, the atomic number increases by 1, and the mass number remains the same. This indicates a beta decay. (iii) \(_{89}^{228}\text{Ac} \rightarrow _{89}^{228}\text{Th}\) In this step, the atomic number remains the same and the mass number remains the same. This indicates a gamma decay.
02

Determine isotope with highest activity

To determine the isotope with the highest activity, we can compare their half-lives. The isotope with the shortest half-life will have the highest activity because it is decaying more quickly. Half-lives given: - Th: 14.0 billion years - Ra: 5.75 years - Ac: 6.25 hours Based on these half-lives, isotope Ac has the highest activity.
03

Determine isotope with lowest activity

To determine the isotope with the lowest activity, we can again compare their half-lives. The isotope with the longest half-life will have the lowest activity because it is decaying less rapidly. Based on the half-lives given, isotope Th has the lowest activity.
04

Find the next isotope in the decay chain

According to the problem, the next step in the decay chain is an alpha emission. In an alpha emission, the atomic number decreases by 2 and the mass number decreases by 4. We will apply this to the current isotope (Th) to determine the next isotope. Current isotope: \(_{89}^{228}\text{Th}\) Applying alpha emission: - Atomic number: 89 - 2 = 87 - Mass number: 228 - 4 = 224 The next isotope in the decay chain is: \(_{87}^{224}\text{Fr}\), francium.

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Most popular questions from this chapter

Which of the following nuclides would you expect to be radioactive: \(_{26}^{58} \mathrm{Fe}, _{27}^{60} \mathrm{Co},\) \(_{41}^{92} \mathrm{Nb},\) mercury-202, radium-226? Justify your choices.

Some watch dials are coated with a phosphor, like ZnS, and a polymer in which some of the \(^{1} \mathrm{H}\) atoms have been replaced by \(^{3} \mathrm{H}\) atoms, tritium. The phosphor emits light when struck by the beta particle from the tritium decay, causing the dials to glow in the dark. The half-life of tritium is 12.3 yr. If the light given off is assumed to be directly proportional to the amount of tritium, by how much will a dial be dimmed in a watch that is 50 yr old?

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?

One of the nuclides in each of the following pairs is radioactive. Predict which is radioactive and which is stable: \((\mathbf{a})_{19}^{39} \mathrm{K}\) and \(_{19}^{40} \mathrm{K},\) \((\mathbf{b})^{209} \mathrm{Bi}\) and \(^{208} \mathrm{Bi}\) \((\mathbf{c})\) nickel-58 and nickel-65.

In 2010, a team of scientists from Russia and the United States reported creation of the first atom of element 117, which is named tennessine, and whose symbol is Ts. The synthesis involved the collision of a target of \(_{97}^{249} \mathrm{Bk}\) with accelerated ions of an isotope which we will denote Q. The product atom, which we will call Z, immediately releases neutrons and forms \(_{97}^{249} \mathrm{Bk} :\) $$_{97}^{249} \mathrm{Bk}+\mathrm{Q} \longrightarrow \mathrm{Z} \longrightarrow_{117 \mathrm{Ts}}^{294 \mathrm{Ts}}+3_{0}^{1} \mathrm{n}$$ (a) What are the identities of isotopes Q and Z? (b) Isotope Q is unusual in that it is very long-lived (its half-life is on the order of 1019 yr) in spite of having an unfavorable neutron-to-proton ratio (Figure 21.1). Can you propose a reason for its unusual stability? (c) Collision of ions of isotope Q with a target was also used to produce the first atoms of livermorium, Lv. The initial product of this collision was \(_{116}^{296} \mathrm{Zn}\). What was the target isotope with which Q collided in this experiment?

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