Chapter 21: Problem 18
A radioactive decay series that begins with 90 Th ends with formation of the stable nuclide \({ }^{208} \mathrm{~Pb} .\) How many alpha-particle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays?
Short Answer
Expert verified
The radioactive decay series that begins with 90 Th and ends with the stable nuclide \({ }^{208} \mathrm{~Pb} \) involves 6 alpha-particle emissions and 4 beta-particle emissions.
Step by step solution
01
Identify initial and final nuclides
Initially, the decay series begins with 90 Th, and the final stable nuclide is \({ }^{208} \mathrm{~Pb} \). Let us denote the number of alpha-particle emissions as "a" and the number of beta-particle emissions as "b". The initial atomic number (Z) of Th is 90, and its mass number (A) is not given. The final atomic number (Z) of Pb is 82, and its mass number (A) is 208.
02
Write equations for changes in atomic and mass numbers
Considering the changes in atomic and mass numbers due to alpha and beta emissions, we can write the following equations:
\( Z_{initial} - 2a + b = Z_{final} \)
\( A_{initial} - 4a = A_{final} \)
03
Calculate the initial mass number
We can calculate the initial mass number (A) of Thorium by using the given final mass number (208) of Lead. From the equation "A_initial - 4a = A_final", we have:
\( A_{initial} - 4a = 208 \)
Since we are not given the initial mass number, we'll assume the known isotope of Thorium, Th-232:
\( 232 - 4a = 208 \)
04
Determine the number of alpha emissions
Rearrange and calculate the number of alpha emissions:
\( 4a = 232 - 208 \)
\( 4a = 24 \)
Therefore, \(a = 6\). There are 6 alpha-particle emissions.
05
Determine the number of beta emissions
Using the equation for changes in atomic number due to alpha and beta emissions ("Z_initial - 2a + b = Z_final"), plug in the given Z values and the calculated number of alpha emissions:
\( 90 - 2(6) + b = 82 \)
Now, solve for the number of beta emissions:
\( 90 - 12 + b = 82 \)
\( b = 82 - 90 + 12 \)
Therefore, \(b = 4\). There are 4 beta-particle emissions.
06
State the results
The radioactive decay series that begins with 90 Th and ends with the formation of the stable nuclide \({ }^{208} \mathrm{~Pb} \) involves 6 alpha-particle emissions and 4 beta-particle emissions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Alpha-Particle Emission
An essential concept in radioactive decay is the alpha-particle emission. This type of emission occurs when an unstable atomic nucleus releases an alpha particle, which consists of two protons and two neutrons. This effectively means it discharges a helium nucleus.
Alpha particles are relatively large and have a charge of +2 due to their protons. As a result, when an alpha-particle emission occurs, the original atom loses:
- 2 protons
- 2 neutrons
Beta-Particle Emission
Beta-particle emission is another significant process in radioactive decay. It involves the transformation of a neutron into a proton inside the nucleus, with the release of a beta particle, commonly an electron, and an antineutrino.
This process increases the atomic number of the element by one, while the mass number remains unchanged:
- Atomic Number increases by 1
- Mass Number remains the same
Decay Series
A decay series is the sequence of radioactive decays that certain isotopes go through to reach a stable end-state. Starting from a radioactive parent isotope, it undergoes a series of emissions, transforming into various elements until reaching a non-radioactive nucleus.
In the decay series from Thorium-232 to Lead-208:
- The series begins with Thorium-232
- Undergoes 6 alpha-particle emissions
- Completes with 4 beta-particle emissions
- Ends as Lead-208, a stable element