Question

Thorium-232 ( \(^{232}\) Th) decays via \(\alpha\) decay. Write out the reaction and identify the daughter nuclide.

Short answer

Answer: The daughter nuclide formed is Radium-228 (\(^{228}\)Ra).

Step-by-step solution

Understanding Alpha Decay

Alpha decay is a radioactive decay process in which an unstable nucleus emits an alpha particle, made of 2 protons and 2 neutrons; so it is a helium-4 nucleus (\(^4_2\)He). This leads to the formation of a new nucleus that has 2 less protons and 2 less neutrons than the original nucleus.

Write the Initial Nuclide

The given nuclide is Thorium-232 (\(^{232}\)Th). This implies that it has 232 nucleons (protons and neutrons), and since it is Thorium, it has an atomic number of 90 (90 protons). Therefore, the initial nuclide can be represented as \(^{232}_{90}\)Th.

Represent the Alpha Particle

As mentioned in Step 1, an alpha particle is equivalent to a helium-4 nucleus, with 2 protons and 2 neutrons. Thus, we can write the alpha particle as \(^4_2\)He.

Write the Alpha Decay Reaction

During alpha decay, the initial nuclide emits the alpha particle. To write the reaction, place the initial nucleus, the alpha particle, and the daughter nucleus side by side, maintaining the conservation of mass number (number of nucleons) and conservation of atomic number (number of protons). The alpha decay reaction would look like: \(^{232}_{90}\)Th \(\rightarrow\) \(^4_2\)He + Daughter Nuclide

Determine the Daughter Nuclide

To determine the daughter nuclide, subtract the mass number and atomic number of the alpha particle from the mass number and atomic number of the initial Thorium-232 nucleus, respectively: Mass number: \(232 - 4 = 228\) Atomic number: \(90 - 2 = 88\) Now, the daughter nucleus has a mass number of 228 and an atomic number of 88, which corresponds to the element Radium (Ra). Therefore, the daughter nuclide is \(^{228}_{88}\)Ra.

Write the Complete Alpha Decay Reaction

Now that we have the daughter nuclide, we can write the complete alpha decay reaction: \(^{232}_{90}\)Th \(\rightarrow\) \(^4_2\)He + \(^{228}_{88}\)Ra The daughter nuclide is Radium-228 (\(^{228}\)Ra).