Problem 18 - Chemistry The Central Science

Question

A radioactive decay series that begins with \(_{90}^{232}\) \(\mathrm{Th}\) ends with formation of the stable nuclide \(_{82}^{208}\) \(\mathrm{Pb}\) . How many alpha-particle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays?

Step-by-Step Solution

Answer

In the radioactive decay series from \(_{90}^{232}\) Th to \(_{82}^{208}\) Pb, there are 4 alpha-particle emissions and no beta-particle emissions involved.

Step 1: Analyze the initial and final isotopes

We begin with the isotope \(_{90}^{232}\) Th and we end with the stable isotope \(_{82}^{208}\) Pb. The changes in atomic number (ΔZ) and mass number (ΔA) during the decay sequence can be expressed as: ΔZ = 90 - 82 = 8 ΔA = 232 - 208 = 24

Step 2: Calculate the number of alpha emissions in the decay series

Each alpha decay reduces the mass number (A) by 4 and the atomic number (Z) by 2. Since the change in atomic number ΔZ is 8 and the change in mass number ΔA is 24, we can write an equation to calculate the number of alpha emissions (n_alpha) involved in the decay sequence: ΔZ = 2 * n_alpha 8 = 2 * n_alpha n_alpha = 4 Therefore, there are 4 alpha-particle emissions in the decay sequence.

Step 3: Calculate the number of beta emissions in the decay series

Each beta decay increases the atomic number (Z) by 1 but does not change the mass number (A). We already determined that the change in atomic number ΔZ is 8. Since we have 4 alpha-particle emissions, which reduce the atomic number by 2 each, we need to calculate the number of beta emissions (n_beta) that compensate this change: ΔZ = 2 * n_alpha - n_beta 8 = 2 * 4 - n_beta 8 = 8 - n_beta n_beta = 0 Therefore, there are no beta-particle emissions involved in this decay sequence.

Step 4: Summarize the results

In this radioactive decay series, there are 4 alpha-particle emissions and no beta-particle emissions involved in the decay sequence from \(_{90}^{232}\) Th to \(_{82}^{208}\) Pb.

Key Concepts

Alpha-Particle Emissions

Alpha particles are composed of 2 protons and 2 neutrons, which is essentially a helium nucleus. When an alpha particle is emitted from a radioactive nucleus, it reduces the atomic number by 2 and the mass number by 4.

This type of decay is quite common in heavy, unstable nuclei, such as uranium and thorium isotopes, which aim to achieve more stability. Therefore, during radioactive decay, when a nuclide emits an alpha particle, it transforms into a different element due to the loss of protons.

In our exercise case,

• the thorium isotope \(_{90}^{232} \mathrm{Th}\) undergoes alpha-particle emissions,
• causing a reduction in the atomic number by 8 and the mass number by 24,
• leading to the stable lead isotope \(_{82}^{208} \mathrm{Pb}\).

This means there are 4 alpha-particle emissions in this decay series.

Beta-Particle Emissions

Beta particles are either electrons or positrons that are emitted from a nucleus. For beta decay, a neutron changes into a proton, thus increasing the atomic number of the element by one. This can occur without a change in the mass number, hence why beta decay often compensates the change in atomic number in decay series without affecting the mass number.

In nuclear reactions, beta decay helps to maintain the balance between protons and neutrons.

In the provided decay series, however, there are no beta-particle emissions:

• The calculation shows a total change in atomic number (ΔZ) is 8, achieved solely through alpha-particle emissions which reduce the atomic number by 2, for a total of 4 reductions.
• This matches with the spacing in protons needed between \(_{90}^{232} \mathrm{Th}\) and \(_{82}^{208} \mathrm{Pb}\).

Thus, in this specific decay series, no beta decays are necessary.

Decay Series Calculations

Understanding how to calculate decay series involves knowing both the alpha and beta particle emissions.

• Alpha emissions decrease the atomic number by 2 and the mass number by 4 each time.
• Beta emissions increase the atomic number by 1 without changing the mass number.

In the exercise, we determined the change in atomic number (ΔZ) was 8 and change in mass number (ΔA) was 24.

With each alpha emission creating a change of ΔZ = 2 and ΔA = 4,

• we found that 4 alpha emissions account for a total reduction in atomic number by 8 and mass number by 16.
• The atomic number was thus reduced solely due to alpha emissions, leading to no requirement for beta emissions.

These calculations are crucial to predicting the final stable element at the end of decay series.