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Chapter 21: Problem 28

In the thorium decay series, thorium- 232 loses a total of \(6 \alpha\) particles and \(4 \beta\) particles in a 10 -stage process. What is the final isotope produced?

Short Answer

Expert verified
The final isotope produced is lead-208 (Pb-208).

Step by step solution

01

Account for Alpha Particles

For every alpha particle (\( \alpha \)) lost in decay, the atomic number decreases by 2 and mass number decreases by 4. Thorium-232 loses 6 alpha particles. Therefore, the atomic number decreases by \(6 * 2 = 12\) and mass number decreases by \(6 * 4 = 24\). The new atomic number after alpha decay is \(90 - 12 = 78\) and new mass number is \(232 - 24 = 208\).
02

Account for Beta Particles

For every beta particle (\( \beta \)) lost in decay, the atomic number increases by 1 and mass number does not change. Thorium-232 loses 4 beta particles. Therefore, the atomic number increases by \(4 * 1 = 4\). The new atomic number after beta decay is \(78 + 4 = 82\). The mass number remains the same because beta decay does not affect it.
03

Identify the Final Isotope

The final isotope has an atomic number of 82 and a mass number of 208. From the periodic table of elements, atomic number 82 is lead (Pb). Therefore, the final isotope produced is lead-208 (Pb-208).

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Most popular questions from this chapter

Consider the decay series \(\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C} \longrightarrow \mathrm{D}\) where \(A, B,\) and \(C\) are radioactive isotopes with halflives of \(4.50 \mathrm{~s}, 15.0\) days, and \(1.00 \mathrm{~s},\) respectively, and \(\mathrm{D}\) is nonradioactive. Starting with 1.00 mole of A, and none of \(\mathrm{B}, \mathrm{C},\) or \(\mathrm{D},\) calculate the number of moles of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) left after 30 days.

A freshly isolated sample of \(90 \mathrm{Y}\) was found to have an activity of \(9.8 \times 10^{5}\) disintegrations per minute at 1: 00 P.M. on December \(3,2000 .\) At 2: 15 P.M. on December \(17,2000,\) its activity was redetermined and found to be \(2.6 \times 10^{4}\) disintegrations per minute. Calculate the half-life of \(90 \mathrm{Y}\).

Tritium, \({ }^{3} \mathrm{H},\) is radioactive and decays by electron emission. Its half-life is 12.5 yr. In ordinary water the ratio of \({ }^{1} \mathrm{H}\) to \({ }^{3} \mathrm{H}\) atoms is \(1.0 \times 10^{17}\) to \(1 .\) (a) Write a balanced nuclear equation for tritium decay. (b) How many disintegrations will be observed per minute in a \(1.00-\mathrm{kg}\) sample of water?

Why do heavy elements such as uranium undergo fission while light elements such as hydrogen and lithium undergo fusion?

Write balanced nuclear equations for these reactions and identify \(\mathrm{X}:\) (a) \(\mathrm{X}(\mathrm{p}, \alpha){ }_{6}^{12} \mathrm{C},\) (b) \({ }_{13}^{27} \mathrm{Al}(\mathrm{d}, \alpha) \mathrm{X}\) (c) \({ }_{25}^{55} \mathrm{Mn}(\mathrm{n}, \gamma) \mathrm{X}\)
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