Problem 26
Question
Rado \(220 \mathrm{Rn}\) produces a daughter nucleus that is radioactive. The daughter, in turn, produces its own radioactive daughter, and so on. This process continues until lead \(83 \underline{208} \mathrm{~b}\) is reached. What are the total number \(N_{\alpha}\) of \(\alpha\) particles and the total number \(N_{\beta}\) of \(\beta\) particles that are generated in this series of radioactive decays?
Step-by-Step Solution
Verified Answer
3 \(\alpha\) particles and 2 \(\beta\) particles are generated.
1Step 1: Identify the Parent and Final Nucleus
The sequence starts with radon-220 \(_{86}^{220} \mathrm{Rn}\) and ends with lead-208 \(_{82}^{208} \mathrm{Pb}\). We need to calculate the changes in atomic and mass numbers to determine the \( \alpha \) and \( \beta \) decays involved.
2Step 2: Determine the Change in Mass Number
The original radon-220 has a mass number of 220, while lead-208 has a mass number of 208. To find the total change, subtract the final mass number from the initial mass number: \( 220 - 208 = 12 \). This indicates how many \( \alpha \) particles, each removing 4 mass units, were emitted:\[ \frac{12}{4} = 3 \].Thus, 3 \( \alpha \) decays occur.
3Step 3: Calculate Total Atomic Number Change
The atomic number changes from 86 in radon-220 to 82 in lead-208. Each \( \alpha \) decay decreases the atomic number by 2. Therefore, with 3 \( \alpha \) decays, the atomic number should drop by \((3 \times 2 = 6) \), reaching 80.
4Step 4: Adjust Atomic Number for Beta Decays
Since the final atomic number is 82 and we need to account for each \( \beta^- \) decay that increases the atomic number by 1, calculate the number of \( \beta \) decays required:\( 82 - 80 = 2 \).Thus, 2 \( \beta \) decays are needed to adjust the atomic number from 80 to 82.
5Step 5: Summarize the Particle Counts
From the analysis:- \( N_{\alpha} = 3 \) - \( N_{\beta} = 2 \)So, the series of decays releases 3 \( \alpha \) particles and 2 \( \beta \) particles.
Key Concepts
Alpha decay in nuclear decay seriesUnderstanding beta decayThe significance of mass numberRole of atomic numberUnderstanding radioactive decay sequences
Alpha decay in nuclear decay series
Alpha decay is a type of radioactive decay where an unstable atom emits an alpha particle, consisting of 2 protons and 2 neutrons. This particle resembles a helium nucleus. When an alpha particle is emitted, the original atom loses 4 mass units and 2 atomic numbers.
In the case of radon-220 decaying to lead-208, alpha decay plays a critical role. Each alpha decay removes components that reduce the mass number and the atomic number of the original atom.
In the case of radon-220 decaying to lead-208, alpha decay plays a critical role. Each alpha decay removes components that reduce the mass number and the atomic number of the original atom.
- Mass number decreases by 4 with each emission.
- Atomic number drops by 2 for each decay event.
Understanding beta decay
Beta decay is another frequent process in nuclear decay. Unlike alpha decay, beta decay involves the transformation of a neutron into a proton within the nucleus, resulting in the emission of a beta particle (electron or positron). This process alters the atomic number without changing the mass number.
During beta decay:
During beta decay:
- The atomic number increases by 1 if a beta-minus particle (electron) is released.
- The mass number remains constant.
The significance of mass number
The mass number of an atom is the total count of protons and neutrons within its nucleus. It is central to understanding nuclear reactions because it indicates how much matter is inside the nucleus.
In decay series problems, observing the mass number helps us understand how many particles like alpha are emitted. Since each alpha decay decreases the mass number by 4, one can trace how the mass number evolves from a parent to a stable daughter nucleus.
In decay series problems, observing the mass number helps us understand how many particles like alpha are emitted. Since each alpha decay decreases the mass number by 4, one can trace how the mass number evolves from a parent to a stable daughter nucleus.
Role of atomic number
The atomic number is key to defining an element's identity. It specifies the number of protons in the nucleus, determining the element's name and position on the periodic table.
In nuclear decay, the atomic number is vital for tracking changes in an element's identity. Alpha decay causes the atomic number to fall by 2, while beta decay increases it by 1. By looking at these shifts, one can map the sequence of transformations that a nucleus undergoes through radioactive decay.
In nuclear decay, the atomic number is vital for tracking changes in an element's identity. Alpha decay causes the atomic number to fall by 2, while beta decay increases it by 1. By looking at these shifts, one can map the sequence of transformations that a nucleus undergoes through radioactive decay.
Understanding radioactive decay sequences
Radioactive decay sequences describe a chain of transformations where an unstable nucleus undergoes a series of emissions to achieve stability. These emissions can be alpha, beta, or other less common types of decay.
This decay sequence for radon-220 to lead-208 is a perfect example, which includes a series of 3 alpha decays and 2 beta decays. Each step in the sequence moves the nucleus closer to a stable configuration:
This decay sequence for radon-220 to lead-208 is a perfect example, which includes a series of 3 alpha decays and 2 beta decays. Each step in the sequence moves the nucleus closer to a stable configuration:
- Alpha decays gradually decrease both the mass and atomic numbers.
- Beta decays adjust the atomic number, fine-tuning the balance of protons and neutrons.
Other exercises in this chapter
Problem 23
Find the energy released when lead \(\underset{82}^{211} \mathrm{~Pb}\) (atomic mass \(=210.988735 \mathrm{u}\) ) undergoes \(\beta^{-}\) decay to become bismut
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Refer to Interactive Solution 31.25 at to review a model for solving this type of problem. Polonium \({ }_{84}^{210} \mathrm{P} \circ(\) atomic mass \(=209.9828
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Find the energy (in MeV) released when \(\beta^{+}\) decay converts sodium \(\underset{11}{22} \mathrm{Na}\) (atomic mass \(=21.994434 \mathrm{u}\) ) into neon
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Interactive LearningWare 31.1 at reviews the concepts that are involved in this problem. An isotope of beryllium (atomic mass \(=7.017 \mathrm{u}\) ) emits a \(
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