Write down equations to describe the alpha decay of the following atoms: a) \(^{212} \mathrm{Rn}\) b) \({ }^{241} \mathrm{Am}\)

An alpha particle is a helium-4 nucleus, which consists of 2 protons and 2 neutrons. Therefore, the alpha particle can be represented as \(^4\mathrm{He}\).

The given parent atoms are: a) \(^{212}\mathrm{Rn}\) b) \(^{241}\mathrm{Am}\)

When a parent atom undergoes alpha decay, it loses 2 protons and 2 neutrons (which are contained in the alpha particle). Therefore, the atomic number (Z) of the daughter atom will be 2 less than the parent atom, and the mass number (A) will be 4 less than the parent atom. a) For \(^{212}\mathrm{Rn}\) (Rn has an atomic number of 86): Daughter atom will have an atomic number of 86 - 2 = 84 Daughter atom will have a mass number of 212 - 4 = 208 So, the daughter atom will be \(^{208}\mathrm{Po}\) (Polonium). b) For \(^{241}\mathrm{Am}\) (Am has an atomic number of 95): Daughter atom will have an atomic number of 95 - 2 = 93 Daughter atom will have a mass number of 241 - 4 = 237 So, the daughter atom will be \(^{237}\mathrm{Np}\) (Neptunium).

Finally, we can write down the equations for the alpha decays: a) \(^{212}\mathrm{Rn} \rightarrow \, ^{208}\mathrm{Po} + ^4\mathrm{He}\) b) \(^{241}\mathrm{Am} \rightarrow \, ^{237}\mathrm{Np} + ^4\mathrm{He}\)