One type of commercial smoke detector contains a minute amount of radioactive americium- \(241\left({ }^{241} \mathrm{Am}\right)\), which decays by \(\alpha\) -particle production. The \(\alpha\) particles ionize molecules in the air, allowing it to conduct an electric current. When smoke particles enter, the conductivity of the air is changed and the alarm buzzes. a. Write the equation for the decay of \({ }^{241} \mathrm{Am}\) by \(\alpha\) -particle production. b. The complete decay of \({ }^{241}\) Am involves successively \(\alpha, \alpha\), \(\beta, \alpha, \alpha, \beta, \alpha, \alpha, \alpha, \beta, \alpha\), and \(\beta\) production. What is the final stable nucleus produced in this decay series? c. Identify the 11 intermediate nuclides.

The element involved in this decay is americium-241 (\(^{241}Am\)). An \(\alpha\) particle is equivalent to a helium nucleus with 2 protons and 2 neutrons, represented as \(^{4}He\).

According to \(\alpha\)-particle decay, the equation can be written as: \( ^{241}Am \rightarrow ^{4}He + ^{237}X \) Here, X represents the resulting nucleus, which has 2 fewer protons and 2 fewer neutrons than \(^{241} Am\). Since \(Am\) has an atomic number of \(\textit{95}\), X will have an atomic number of \(\textit{93}\). The element with an atomic number 93 is Neptunium (Np). So, the decay equation is: \( ^{241}Am \rightarrow ^{4}He + ^{237}Np \) b. What is the final stable nucleus produced in this decay series?

The decay series includes 11 successive decays: \(\alpha, \alpha, \beta, \alpha, \alpha, \beta, \alpha, \alpha, \alpha, \beta, \alpha,\) and \(\beta\). For each \(\alpha\) decay, the atomic number decreases by 2 and the mass number decreases by 4. For each \(\beta\) decay, the atomic number increases by 1 and the mass number remains unchanged. In order to find the final stable nucleus, we need to account for these changes in atomic number and mass number. We will first count the total number of alpha and beta decays: There are 8 alpha decays and 4 beta decays. Now, let's find the changes in atomic number and mass number due to these decays: - Due to alpha decays: \(\Delta Z_\alpha = -2 \times 8 = -16\) \(\Delta A_\alpha = -4 \times 8 = -32\) - Due to beta decays: \(\Delta Z_\beta = 1 \times 4 = 4\) \(\Delta A_\beta = 0\) Now, adding these changes to the initial atomic number and mass number of \(^{241}Am\): \(Z_{final} = Z_{Am} + (\Delta Z_\alpha + \Delta Z_\beta) = 95 - 16 + 4 = 83\) \(A_{final} = A_{Am} + (\Delta A_\alpha + \Delta A_\beta) = 241 - 32 = 209\) The final stable nucleus has an atomic number of 83 and a mass number of 209, which corresponds to Bismuth-209 (\(^{209}Bi\)). c. Identify the 11 intermediate nuclides.

We calculate each intermediate nuclide by following the decay series and the corresponding changes in atomic number and mass number: 1. \(^{241}Am \xrightarrow{\alpha} ^{237}Np\) 2. \(^{237}Np \xrightarrow{\alpha} ^{233}U\) 3. \(^{233}U \xrightarrow{\beta} ^{233}Np\) 4. \(^{233}Np \xrightarrow{\alpha} ^{229}Pa\) 5. \(^{229}Pa \xrightarrow{\alpha} ^{225}Ac\) 6. \(^{225}Ac \xrightarrow{\beta} ^{225}Th\) 7. \(^{225}Th \xrightarrow{\alpha} ^{221}Ra\) 8. \(^{221}Ra \xrightarrow{\alpha} ^{217}Fr\) 9. \(^{217}Fr \xrightarrow{\alpha} ^{213}At\) 10. \(^{213}At \xrightarrow{\beta} ^{213}Bi\) 11. \(^{213}Bi \xrightarrow{\alpha} ^{209}Tl\) 12. \(^{209}Tl \xrightarrow{\beta} ^{209}Pb\) 13. \(^{209}Pb \xrightarrow{\beta} ^{209}Bi\) The 11 intermediate nuclides in the decay series are: \(^{237}Np\), \(^{233}U\), \(^{233}Np\), \(^{229}Pa\), \(^{225}Ac\), \(^{225}Th\), \(^{221}Ra\), \(^{217}Fr\), \(^{213}At\), \(^{213}Bi\), and \(^{209}Tl\). The final stable nucleus is \(^{209}Bi\).