Question

A radioactive decay series that begins with \({ }_{90}^{232}\) Th ends with formation of the stable nuclide \({ }_{82}^{208} \mathrm{~Pb}\). How many alphaparticle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays?

Short answer

In the radioactive decay series starting with \({ }_{90}^{232}\) Th and ending with \({ }_{82}^{208}\mathrm{Pb}\), there are 6 alpha-particle emissions and 20 beta-particle emissions.

Step-by-step solution

Calculate the change in atomic number and atomic mass

To find how many alpha and beta emissions occurred, we need to calculate the difference in atomic number and atomic mass between the initial and final nuclides. Initial nuclide: \({ }_{90}^{232}\) Th Final nuclide: \({ }_{82}^{208}\mathrm{Pb}\) Change in atomic number: \(\Delta Z = 90 - 82 = 8\) Change in atomic mass: \(\Delta A = 232 - 208 = 24\)

Find the number of alpha decays

Since each alpha decay reduces the atomic mass by 4, we can calculate the number of alpha decays by dividing the total change in atomic mass by 4: Number of alpha decays = \(\frac{\Delta A}{4} = \frac{24}{4} = 6\)

Find the number of beta decays

Since each alpha decay reduces the atomic number by 2 and each beta decay increases the atomic number by 1, we can calculate the number of beta decays by using the fact that the total number of alpha and beta decays results in the observed change in atomic number. Let \(a\) be the number of alpha decays and \(b\) be the number of beta decays, then: \(-2a + b = \Delta Z\) We already found the number of alpha decays (\(a = 6\)), so we can plug that back into the equation: \(-2(6) + b = 8\) Solving for \(b\): \(b = 8 + 2(6) = 20\)

Conclusion

In the radioactive decay series starting with \({ }_{90}^{232}\) Th and ending with \({ }_{82}^{208}\mathrm{Pb}\), there are 6 alpha-particle emissions and 20 beta-particle emissions.