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In the thorium decay series, thorium- 232 loses a total of six \(\alpha\) particles and four \(\beta\) particles in a 10 -stage process. What is the final isotope produced?

Short Answer

Expert verified
The final isotope produced is lead-208 (\(^{208}\text{Pb}\)).

Step by step solution

01

Understand Alpha and Beta Decay

In an \(\alpha\) decay, the nucleus loses two protons and two neutrons, which changes the element into another element with an atomic number decreased by 2.In a \(\beta\) decay, a neutron in the nucleus is converted into a proton, increasing the atomic number by 1 without changing the mass number.
02

Calculate Change from Alpha Decay

Given that thorium-232 loses 6 \(\alpha\) particles: - Each \(\alpha\) decay decreases the atomic number by 2. So, total decrease in atomic number is \(6 \times 2 = 12\).- Each \(\alpha\) decay decreases the mass number by 4. So, the total decrease in mass number is \(6 \times 4 = 24\).
03

Calculate Change from Beta Decay

Given that thorium-232 undergoes 4 \(\beta\) decays:- Each \(\beta\) decay increases the atomic number by 1. So, total increase in atomic number is \(4 \times 1 = 4\).- \(\beta\) decay does not change the mass number.
04

Calculate Final Atomic and Mass Numbers

Start with thorium-232, which has an atomic number of 90:- From \(\alpha\) decay: Atomic number decreases by 12.- From \(\beta\) decay: Atomic number increases by 4.- Final atomic number: \(90 - 12 + 4 = 82\).- From \(\alpha\) decay: Mass number decreases by 24.- Initial mass number is 232, so final mass number is \(232 - 24 = 208\).
05

Identify the Isotope

The element with atomic number 82 is lead (Pb). Therefore, after the decay process, the final isotope produced is lead-208 (\(^{208}\text{Pb}\)).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alpha Decay
Alpha decay is a type of radioactive decay where an unstable atomic nucleus releases an alpha particle. An alpha particle consists of two protons and two neutrons, which is the equivalent of a helium nucleus. This release causes the original atom's nucleus to lose mass and reduce its atomic number. This is because it sheds those two protons, changing into a completely different element. For instance, when thorium-232 undergoes alpha decay, it loses two protons, changing into the element with an atomic number that is 2 less than thorium's. This process results in a decrease in atomic number by 2 and in mass number by 4 for each alpha decay event.
  • Alpha particles are relatively massive in comparison to other nuclear decay particles like beta particles.
  • Because of their mass, alpha particles do not penetrate materials deeply and can be stopped by a sheet of paper or even skin.
  • Alpha decay is typical in heavy elements like uranium, radium, or thorium.
Beta Decay
Beta decay involves the transformation of a neutron into a proton within the nucleus, accompanied by the emission of a beta particle. A beta particle is essentially a high-speed electron or positron. Unlike alpha decay, beta decay results in an increase in the atomic number because the neutron changes into a proton, adding an extra positive charge to the nucleus. This conversion does not affect the mass number because the emitted electron or positron has negligible mass compared to nucleons.
  • The process of beta decay includes the emission of an electron neutrino or antineutrino alongside the beta particle.
  • Beta particles are lighter and can penetrate materials further than alpha particles, usually requiring materials like aluminum to be stopped.
  • This type of decay is common in isotopes that are neutron-rich.
Thorium Decay Series
The thorium decay series is a chain of radioactive decays beginning with thorium-232 and ending with a stable isotope of lead, lead-208. This series involves multiple steps, with the key stages being six alpha decays and four beta decays. In the thorium decay series, each of these decay processes alters the element's identity and advances it towards a stable state. Thorium-232, through a series of alpha and beta decays, typically changes into elements that are progressively lighter, eventually reaching lead-208.
  • The thorium decay series spans across 10 stages, demonstrating complex transformation of the parent isotope.
  • The physical and chemical characteristics of intermediate isotopes vary, showcasing diverse changes in their stability and radioactive properties.
  • The journey from thorium to lead illustrates the intricate dance of nuclear transformations occurring over time, highlighting the natural processes that stabilize atomic nuclei.
Isotope Identification
Isotope identification in the context of nuclear chemistry is about determining the final result of a series of decay processes based on changes in atomic and mass numbers. In the thorium decay series, these changes guide us to identify which isotope is eventually formed. By understanding both alpha and beta decay processes: alpha decay reduces the atomic number by 2, and beta decay increases it by 1, we can predict the end state of a nuclear decay sequence. For thorium-232, after undergoing these decay processes, it results in lead-208. This involves considering initial and final atomic and mass numbers, following formula calculations akin to:
  • Final Atomic Number = Initial Atomic Number - (2 * Number of Alpha Decays) + (1 * Number of Beta Decays)
  • Final Mass Number = Initial Mass Number - (4 * Number of Alpha Decays)
By calculating correctly, the identification of the final isotope becomes straightforward. The final atomic number points to the final element on the periodic table, in this case, lead (Pb).

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Most popular questions from this chapter

Explain how you might use a radioactive tracer to show that ions are not completely motionless in crystals.

Nuclear waste disposal is one of the major concerns of the nuclear industry. In choosing a safe and stable environment to store nuclear wastes, consideration must be given to the heat released during nuclear decay. As an example, consider the \(\beta\) decay of \({ }^{90} \mathrm{Sr}(89.907738 \mathrm{amu})\) : $$ { }_{38}^{90} \mathrm{Sr} \longrightarrow{ }_{39}^{90} \mathrm{Y}+{ }_{-1}^{0} \beta \quad t_{1 / 2}=28.1 \mathrm{yr} $$ The \({ }^{90} \mathrm{Y}(89.907152 \mathrm{amu})\) further decays as follows: \({ }_{39}^{90} \mathrm{Y} \longrightarrow{ }_{40}^{90} \mathrm{Zr}+{ }_{-1}^{0} \beta \quad t_{1 / 2}=64 \mathrm{~h}\) Zirconium- \(90(89.904703 \mathrm{amu})\) is a stable isotope. (a) Use the mass defect to calculate the energy released (in joules) in each of the preceding two decays. (The mass of the electron is \(5.4857 \times 10^{-4}\) amu.) (b) Starting with 1 mole of \({ }^{90} \mathrm{Sr}\), calculate the number of moles of \({ }^{90} \mathrm{Sr}\) that will decay in a year. (c) Calculate the amount of heat released (in kJ) corresponding to the number of moles of \({ }^{90} \mathrm{Sr}\) decayed to \({ }^{90} \mathrm{Zr}\) in part \((\mathrm{b}).\)

What are the advantages of a fusion reactor over a fission reactor? What are the practical difficulties in operating a large-scale fusion reactor?

For each pair of isotopes listed, predict which one is less stable: (a) \({ }_{3}^{6} \mathrm{Li}\) or \({ }_{3}^{9} \mathrm{Li}\) (b) \({ }_{11}^{23} \mathrm{Na}\) or \({ }_{11}^{25} \mathrm{Na}\) (c) \({ }_{20}^{48} \mathrm{Ca}\) or \({ }_{21}^{48} \mathrm{Sc}\)

Strontium-90 is one of the products of the fission of uranium-235. This strontium isotope is radioactive, with a half-life of 28.1 years. Calculate how long (in years) it will take for \(1.00 \mathrm{~g}\) of the isotope to be reduced to \(0.200 \mathrm{~g}\) by decay.

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