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Magazine Chapter 8: Problem 18
The \(Q\) value of the decay \({ }^{233} \mathrm{U} \rightarrow{ }^{229} \mathrm{Th}+\alpha\) is \(4.909 \mathrm{MeV}\). Excited states of \({ }^{229} \mathrm{Th}\) at \(29,42.72\), and \(97 \mathrm{keV}\) are populated in the decay. Compute the energies of the five most energetic \(\alpha\) groups emitted in the \({ }^{233} \mathrm{U}\) decav.
Short Answer
Expert verified
4.909 MeV, 4.880 MeV, 4.86628 MeV, 4.812 MeV, 4.812 MeV.
Step by step solution
01
- Understand the problem
The problem involves determining the energies of the five most energetic alpha particles emitted during the decay of \({ }^{233} \text{U} \rightarrow { }^{229} \text{Th} + \text{伪}\), where the Q value (total energy released) is given as 4.909 MeV. Additionally, some excited states of \({ }^{229} \text{Th}\) at \(29 \text{keV}\), \(42.72 \text{keV}\), and \(97 \text{keV}\) are populated.
02
- Calculate the ground state alpha energy
When \({ }^{233} \text{U}\) decays and the daughter nucleus \({ }^{229} \text{Th}\) is in its ground state (no excitation), the energy of the emitted alpha particle is equal to the Q value:e_{\text{伪, ground state}} = 4.909 \text{MeV}.
03
- Calculate alpha energy for Th excited at 29 keV
If \({ }^{229} \text{Th}\) is in the \(29 \text{keV}\) excited state, then the alpha particle's energy will be reduced by \(29 \text{keV}\) (0.029 MeV):e_{\text{伪, 29 keV}} = 4.909 \text{MeV} - 0.029 \text{MeV} = 4.880 \text{MeV}.
04
- Calculate alpha energy for Th excited at 42.72 keV
If \({ }^{229} \text{Th}\) is in the \(42.72 \text{keV}\) excited state, then the alpha particle's energy will be reduced by \(42.72 \text{keV}\) (0.04272 MeV):e_{\text{伪, 42.72 keV}} = 4.909 \text{MeV} - 0.04272 \text{MeV} = 4.86628 \text{MeV}.
05
- Calculate alpha energy for Th excited at 97 keV
If \({ }^{229} \text{Th}\) is in the \(97 \text{keV}\) excited state, then the alpha particle's energy will be reduced by \(97 \text{keV}\) (0.097 MeV):e_{\text{伪, 97 keV}} = 4.909 \text{MeV} - 0.097 \text{MeV} = 4.812 \text{MeV}.
06
- Summarize the results
The five most energetic alpha groups correspond to the decay with the three different excited states, plus any repetitions or combinations not specified in the problem. So the energies are: 4.909 MeV, 4.880 MeV, 4.86628 MeV, 4.812 MeV. For the fifth energy, we will take the next possible excited state, which we don鈥檛 have explicit information for, hence reconsidering the least energetic state twice: 4.812 MeV.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Decay Energies
In nuclear physics, decay energy is vital to understand how particles interact and transform. The decay energy, also known as the Q value, represents the total energy released during a nuclear decay process. In the given problem, the decay of \(^{233}\text{U}\) into \(^{229}\text{Th}\) and an alpha particle (\text{伪}), the Q value is \(4.909\text{MeV}\). Calculating the energy of emitted particles helps understand the decay mechanism and resulting particle states. These concepts apply widely, from understanding natural radioactivity to applications in nuclear medicine.
Alpha Particles
Alpha particles are crucial in nuclear decay. They consist of two protons and two neutrons, making them the same as a helium-4 nucleus. Alpha particles are emitted in many types of radioactive decay, including the decay of uranium-233 described in the exercise. They are relatively heavy and positively charged. This makes them interact strongly with matter, losing energy quickly and having a short range. Understanding alpha particles is important in nuclear physics, radiation safety, and detecting radioactive materials.
Q Value
The Q value is central in nuclear decay calculations. It signifies the total energy released in the decay process. For the decay \(^{233}\text{U} \rightarrow ^{229}\text{Th}+\alpha\), the Q value is given as \(4.909\text{MeV}\). This Q value is crucial because it delineates the maximum possible kinetic energy of the emitted particles. Understanding it is the first step in dissecting how this energy is distributed among the particles, and how daughter nuclei might populate various excited states.
Excited States
In nuclear physics, excited states refer to a higher energy level of a nucleus compared to its ground state. When the decay of \(^{233}\text{U}\) produces \(^{229}\text{Th}\), it may leave \(^{229}\text{Th}\) in an excited state. The given problem lists excited states at \(29\text{keV}\), \(42.72\text{keV}\), and \(97\text{keV}\). When a nucleus decays to an excited state, the energy of the emitted alpha particle is decreased by the energy of the excited state. Hence, understanding these states helps in calculating the exact energy of the alpha particles.
Energy Calculation Steps
Energy calculation follows clearly defined steps:
- Identify the Q value of the decay.
- Calculate the alpha particle energy assuming no excitation, which here is the Q value itself, \(4.909\text{MeV}\).
- Subtract the energy of the excited state from the Q value to find the alpha particle's energy for each excited state:
- For \(29\text{keV}\) excitation: \(4.880\text{MeV}\).
- For \(42.72\text{keV}\) excitation: \(4.86628\text{MeV}\).
- For \(97\text{keV}\) excitation: \(4.812\text{MeV}\).聽
- Identify the five most energetic alpha groups. Given the problem conditions, they are \(4.909\text{MeV}\), \(4.880\text{MeV}\), \(4.86628\text{MeV}\), \(4.812\text{ MeV}\), and again \(4.812\text{MeV}\).
