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Mobile App Chapter 10: Problem 31
(a) Calculate BEN for \(^{235} \mathrm{U}\), the rarer of the two most common uranium isotopes; (b) Calculate BEN for \(^{238} \mathrm{U}\) (Most of uranium is \(^{238} \mathrm{U}\).)
Short Answer
Expert verified
The binding energy per nucleon for \(^{235} \mathrm{U}\) is approximately 3.146 MeV/nucleon, and for \(^{238} \mathrm{U}\) is approximately 3.136 MeV/nucleon.
Step by step solution
01
Determine the number of nucleons in each isotope
We first determine the number of nucleons, which are protons and neutrons, in each isotope. The number of nucleons is the same as the mass number of the isotope:
- For \(^{235} \mathrm{U}\), there are 235 nucleons
- For \(^{238} \mathrm{U}\), there are 238 nucleons
02
Look up the masses of the isotopes
We need to find the masses of the uranium isotopes in atomic mass units (u). From a table of nuclear masses, we find:
- Mass of \(^{235} \mathrm{U}\): 235.0439299 u
- Mass of \(^{238} \mathrm{U}\): 238.0507882 u
03
Calculate the mass defect
Next, we calculate the mass defect (∆m) for each isotope. Mass defect is the difference between the mass of the individual nucleons (protons and neutrons) and the actual mass of the isotope:
∆m for \(^{235} \mathrm{U}\) = (number of protons × mass of proton + number of neutrons × mass of neutron) - mass of \(^{235} \mathrm{U}\)
∆m for \(^{238} \mathrm{U}\) = (number of protons × mass of proton + number of neutrons × mass of neutron) - mass of \(^{238} \mathrm{U}\)
But first, we need to find the number of protons and neutrons in each isotope. Since uranium is element 92, it has 92 protons. The mass number minus the number of protons gives us the number of neutrons:
- Number of neutrons in \(^{235} \mathrm{U}\): 235 - 92 = 143
- Number of neutrons in \(^{238} \mathrm{U}\): 238 - 92 = 146
Now we can calculate the mass defects. Using the values for the mass of the proton (1.007276466812 u) and the mass of the neutron (1.008664915 u):
∆m for \(^{235} \mathrm{U}\) = (92 × 1.007276466812 + 143 × 1.008664915) - 235.0439299 = 0.793471 u
∆m for \(^{238} \mathrm{U}\) = (92 × 1.007276466812 + 146 × 1.008664915) - 238.0507882 = 0.799981 u
04
Calculate the binding energy for each isotope
We now use the mass defects to calculate the binding energy (BE) for each isotope, using the formula \(E=mc^2\). We will use the conversion factor 1 u = 931.494095 MeV/c² to change the mass defects from atomic mass units to MeV (Mega-electron volts):
BE for \(^{235} \mathrm{U}\) = 0.793471 u × 931.494095 MeV/c²/u = 739.315 MeV
BE for \(^{238} \mathrm{U}\) = 0.799981 u × 931.494095 MeV/c²/u = 745.991 MeV
05
Calculate the binding energy per nucleon (BEN) for each isotope
Finally, we can calculate the binding energy per nucleon (BEN) by dividing the binding energy by the number of nucleons:
BEN for \(^{235} \mathrm{U}\) = 739.315 MeV / 235 = 3.146 MeV/nucleon
BEN for \(^{238} \mathrm{U}\) = 745.991 MeV / 238 = 3.136 MeV/nucleon
So the binding energy per nucleon for \(^{235} \mathrm{U}\) is approximately 3.146 MeV/nucleon, and for \(^{238} \mathrm{U}\) is approximately 3.136 MeV/nucleon.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nuclear Physics
Nuclear physics is a field of study focused on understanding the components and interactions within the atomic nucleus. This discipline explores various phenomena such as the forces keeping the nuclear particles bound together and the principles behind nuclear reactions, including fission and fusion.
At the core of nuclear physics is the concept of the atom, composed of a dense nucleus surrounded by orbiting electrons. The nucleus itself contains protons with a positive charge and neutrons that are neutral, collectively referred to as nucleons. The numbers of these nucleons define an element’s identity and its isotopes. The stability and energy dynamics of these nucleons are central to nuclear physics and have implications in fields like energy generation, medical treatments, and astrophysics.
At the core of nuclear physics is the concept of the atom, composed of a dense nucleus surrounded by orbiting electrons. The nucleus itself contains protons with a positive charge and neutrons that are neutral, collectively referred to as nucleons. The numbers of these nucleons define an element’s identity and its isotopes. The stability and energy dynamics of these nucleons are central to nuclear physics and have implications in fields like energy generation, medical treatments, and astrophysics.
Uranium Isotopes
Uranium isotopes are variants of uranium atoms that have the same number of protons but a different number of neutrons. The two most common isotopes of uranium are uranium-235 ([^{235}U]) and uranium-238 ([^{238}U]). While both isotopes are naturally occurring, [^{238}U] is more abundant, making up about 99.3% of natural uranium, whereas [^{235}U] accounts for only about 0.7%.
The differing numbers of neutrons in these isotopes lead to variations in their physical and nuclear properties, such as their stability and how they interact with neutrons. This has profound implications for their use in nuclear reactors and weapons. For instance, [^{235}U] is fissile, meaning it can sustain a nuclear chain reaction, which is crucial for both nuclear power plants and atomic bombs.
The differing numbers of neutrons in these isotopes lead to variations in their physical and nuclear properties, such as their stability and how they interact with neutrons. This has profound implications for their use in nuclear reactors and weapons. For instance, [^{235}U] is fissile, meaning it can sustain a nuclear chain reaction, which is crucial for both nuclear power plants and atomic bombs.
Mass Defect
The mass defect is the difference between the predicted mass of an atomic nucleus based on its constituent nucleons and its actual mass. It is a fundamental concept in nuclear physics that results from the binding energy, which is the energy needed to keep the nucleus intact.
This energy is released when a nucleus is formed from protons and neutrons, and it is manifested as a mass defect, according to Einstein's famous equation [E=mc^2], where [E] is energy, [m] is mass, and [c] is the speed of light. A larger mass defect indicates a higher binding energy, which generally means a more stable nucleus. When calculating the mass defect for uranium isotopes, as the example in the problem shows, it helps to understand the stability and potential energy that can be released from these isotopes, which is tremendously important for nuclear power and weapons technology.
This energy is released when a nucleus is formed from protons and neutrons, and it is manifested as a mass defect, according to Einstein's famous equation [E=mc^2], where [E] is energy, [m] is mass, and [c] is the speed of light. A larger mass defect indicates a higher binding energy, which generally means a more stable nucleus. When calculating the mass defect for uranium isotopes, as the example in the problem shows, it helps to understand the stability and potential energy that can be released from these isotopes, which is tremendously important for nuclear power and weapons technology.
