Final answer:
The nuclear equation for the alpha decay of thorium-232 results in the emission of an alpha particle and the production of radium-228, which is represented as ¹°¹Th²³² → ⁴He + ¹°¹Th²²¸.
Step-by-step explanation:
The nuclear equation for the alpha decay of thorium-232 involves the release of an alpha particle, which is a helium nucleus (He) with two protons and two neutrons. The alpha decay will result in a decrease in both atomic number and atomic weight of the original atom. Consequently, thorium-232 (Th²³²) will emit an alpha particle and transform into a different element with an atomic number that is reduced by two and a mass number reduced by four. This gives us:
²³²Th → &sup4;He + ²²&sup8;Ra
To check the balance of the equation, the sum of the mass numbers and atomic numbers on each side must be equal. Thorium-232 has a mass number of 232 and an atomic number of 90. After emitting an alpha particle (He-4), the resulting radium (Ra) will have an atomic number decreased by 2, giving it an atomic number of 88 and a mass number of 228. The correct nuclear equation is thus option (a).