233k views
3 votes
What is the nuclear equation for the alpha decay of thorium-232?

a) ( text{Th}²³² rightarrow text{He}⁴ + text{Ra}²²⁸ )
b) ( text{Th}²³² rightarrow text{He}⁴ + text{Pa}²²⁸ )
c) ( text{Th}²³² rightarrow text{He}⁴ + text{Ac}²²⁸ )
d) ( text{Th}²³² rightarrow text{He}⁴ + text{U}²²⁸ )

User Sudo Work
by
7.0k points

1 Answer

3 votes

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).

User Brabster
by
8.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.