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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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