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 \(^{241}_{95} \mathrm{Am}\) by \(\alpha\) -particle production. b. The complete decay of \(^{241} \mathrm{Am}\) involves successively \(\alpha, \alpha\) \(\boldsymbol{\beta}, \alpha, \alpha, \boldsymbol{\beta}, \alpha, \alpha, \alpha, \boldsymbol{\beta}, \alpha,\) and \(\boldsymbol{\beta}\) production. What is the final stable nucleus produced in this decay series? c. Identify the 11 intermediate nuclides.

To write the equation for the decay of \(\left.^{241}_{95} \mathrm{Am}\right.\) by alpha-particle production, we need to know that an alpha particle is a helium nucleus, represented as \(\left.^{4}_{2} \mathrm{He}\right.\). The decay process is represented as: \[ ^{241}_{95}\mathrm{Am}\rightarrow ^{A}_{Z}\mathrm{X} + ^{4}_{2}\mathrm{He} \] Where A and Z are the atomic mass and atomic number of the resulting nucleus (X), respectively. To find the values for A and Z, use the conservation of nucleons (protons and neutrons): A = 241 - 4 Z = 95 - 2 #(Step 2) Final Stable Nucleus in Decay Series#

The decay series given is: \(\alpha, \alpha, \beta, \alpha, \alpha, \beta, \alpha, \alpha, \alpha, \beta, \alpha, \beta\). Using this decay series, we can calculate the change in atomic mass and atomic number after each step: 1. After the first two alpha decays, Am loses 2 alpha particles, so the change in atomic mass number is -8 and the change in the atomic number is -4. 2. Following the beta decay, there is no change in the atomic mass but an increase in the atomic number by 1. 3. As we proceed through the decay series, the total change in atomic mass and atomic number after all these decays will be -32 and -14, respectively. Now, to find the final stable nucleus: Final Atomic Mass = 241 - 32 = 209 Final Atomic Number = 95 - 14 = 81 The final stable nucleus is \(\left.^{209}_{81}\mathrm{Tl}\right.\). #(Step 3) Identify Intermediate Nuclides#

To find the intermediate nuclides, we will follow the same decay series and find the resulting nuclei after each decay. Nuclide 1: After the first alpha decay, Atomic Mass = 241 - 4 = 237 Atomic Number = 95 - 2 = 93 The first intermediate nuclide is \(\left.^{237}_{93}\mathrm{Np}\right.\). Nuclide 2: After the second alpha decay, Atomic Mass = 237 - 4 = 233 Atomic Number = 93 - 2 = 91 The second intermediate nuclide is \(\left.^{233}_{91}\mathrm{Pa}\right.\). Nuclide 3: After the beta decay, Atomic Mass = 233 Atomic Number = 91 + 1 = 92 The third intermediate nuclide is \(\left.^{233}_{92}\mathrm{U}\right.\). Follow this process for the rest of the decay series to find all 11 intermediate nuclides. The complete list will be: 1. \(\left.^{237}_{93}\mathrm{Np}\right.\) 2. \(\left.^{233}_{91}\mathrm{Pa}\right.\) 3. \(\left.^{233}_{92}\mathrm{U}\right.\) 4. \(\left.^{229}_{90}\mathrm{Th}\right.\) 5. \(\left.^{225}_{88}\mathrm{Ra}\right.\) 6. \(\left.^{225}_{89}\mathrm{Ac}\right.\) 7. \(\left.^{221}_{87}\mathrm{Fr}\right.\) 8. \(\left.^{217}_{85}\mathrm{At}\right.\) 9. \(\left.^{213}_{83}\mathrm{Bi}\right.\) 10. \(\left.^{213}_{84}\mathrm{Po}\right.\) 11. \(\left.^{209}_{82}\mathrm{Pb}\right.\).