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 \({ }_{95}^{241} \mathrm{Am}\) 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.

For an α-decay, the parent nucleus loses an alpha particle (a helium nucleus) with mass number 4 and atomic number 2. So, the mass number decreases by 4 and the atomic number decreases by 2. Let's write the decay equation for Am-241: \[^{241}_{95}\text{Am} \rightarrow ^{237}_{93}\text{X} + ^{4}_{2}\text{He}\] Here, X represents the element with atomic number 93, which is Neptunium (Np). So, the decay equation becomes: \[^{241}_{95}\text{Am} \rightarrow ^{237}_{93}\text{Np} + ^{4}_{2}\text{He}\]

The decay series is given as: α, α, β, α, α, β, α, α, α, β, α, β. We will find the final stable nucleus by successively applying each decay process to Am-241. For each α-decay, the mass number decreases by 4 and the atomic number decreases by 2. For each β-decay, the mass number remains the same, but the atomic number increases by 1. Starting with Am-241, let's apply the given decay series: \[^{241}_{95}\text{Am} \xrightarrow[\alpha]{} ^{237}_{93}\text{Np} \xrightarrow[\alpha]{} ^{233}_{91}\text{Pa} \xrightarrow[\beta]{} ^{233}_{92}\text{U} \xrightarrow[\alpha]{} ^{229}_{90}\text{Th}\] \[^{229}_{90}\text{Th} \xrightarrow[\alpha]{} ^{225}_{88}\text{Ra} \xrightarrow[\beta]{} ^{225}_{89}\text{Ac} \xrightarrow[\alpha]{} ^{221}_{87}\text{Fr}\] \[^{221}_{87}\text{Fr} \xrightarrow[\alpha]{} ^{217}_{85}\text{At} \xrightarrow[\alpha]{} ^{213}_{83}\text{Bi} \xrightarrow[\beta]{} ^{213}_{84}\text{Po}\] \[^{213}_{84}\text{Po} \xrightarrow[\alpha]{} ^{209}_{82}\text{Pb} \xrightarrow[\beta]{} ^{209}_{83}\text{Bi}\] The final stable nucleus in the decay series is Bismuth-209 (Bi-209).

In the above step, we already identified the intermediate nuclides while applying each decay process. The 11 intermediate nuclides in the decay series are: 1. Neptunium-237 (Np-237) 2. Protactinium-233 (Pa-233) 3. Uranium-233 (U-233) 4. Thorium-229 (Th-229) 5. Radium-225 (Ra-225) 6. Actinium-225 (Ac-225) 7. Francium-221 (Fr-221) 8. Astatine-217 (At-217) 9. Bismuth-213 (Bi-213) 10. Polonium-213 (Po-213) 11. Lead-209 (Pb-209)