College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![### Nuclear Reactions: Practice Problem
#### Question 5:
**Complete the following nuclear reaction and identify the missing element (use periodic table)**
(i)
\[ \phantom{^{238}_{92}}\text{U} \rightarrow \phantom{^{4}_{2}}\text{He} + \text{X} \]
Assume the initial isotope is Uranium-238 and it decays, producing an alpha particle (denoted by \(\phantom{^{4}_{2}}\text{He}\)) and another element (denoted by \(\text{X}\)).
### Instructions:
1. Identify the alpha particle: \( \phantom{^{4}_{2}} \text{He} \). It has 2 protons and an atomic mass of 4.
2. Use the conservation of mass and atomic numbers to find the missing element, \(\text{X}\):
- Initial isotope Uranium-238: \( \phantom{^{238}_{92}} \text{U} \) has an atomic number of 92 and a mass number of 238.
- The decay process produces an alpha particle and another nucleus, so the total mass and atomic numbers should be conserved.
Using these details, calculate the atomic number and mass number of the missing element \(\text{X}\).
### Calculation Process:
1. **Mass Number Calculation:**
\[ 238 = 4 + (mass \, number \, of \, X) \]
Therefore,
\[ mass \, number \, of \, X = 238 - 4 = 234 \]
2. **Atomic Number Calculation:**
\[ 92 = 2 + (atomic \, number \, of \, X) \]
Therefore,
\[ atomic \, number \, of \, X = 92 - 2 = 90 \]
Hence, the missing element \(\text{X}\) has an atomic number of 90, which corresponds to Thorium (Th) in the periodic table.
### Completed Reaction:
\[ \phantom{^{238}_{92}}\text{U} \rightarrow \phantom{^{4}_{2}}\text{He} + \phantom{^{234}_{90}}\text{Th} \]
This example shows how the conservation laws for mass and atomic number can help solve for unknowns in nuclear reactions.](https://content.bartleby.com/qna-images/question/8a89208d-bdc5-4565-a1f9-416479803e3f/da74c85f-313a-4b91-a4bf-35b8bfb84e7e/zc2i6uo.jpeg)
Transcribed Image Text:### Nuclear Reactions: Practice Problem
#### Question 5:
**Complete the following nuclear reaction and identify the missing element (use periodic table)**
(i)
\[ \phantom{^{238}_{92}}\text{U} \rightarrow \phantom{^{4}_{2}}\text{He} + \text{X} \]
Assume the initial isotope is Uranium-238 and it decays, producing an alpha particle (denoted by \(\phantom{^{4}_{2}}\text{He}\)) and another element (denoted by \(\text{X}\)).
### Instructions:
1. Identify the alpha particle: \( \phantom{^{4}_{2}} \text{He} \). It has 2 protons and an atomic mass of 4.
2. Use the conservation of mass and atomic numbers to find the missing element, \(\text{X}\):
- Initial isotope Uranium-238: \( \phantom{^{238}_{92}} \text{U} \) has an atomic number of 92 and a mass number of 238.
- The decay process produces an alpha particle and another nucleus, so the total mass and atomic numbers should be conserved.
Using these details, calculate the atomic number and mass number of the missing element \(\text{X}\).
### Calculation Process:
1. **Mass Number Calculation:**
\[ 238 = 4 + (mass \, number \, of \, X) \]
Therefore,
\[ mass \, number \, of \, X = 238 - 4 = 234 \]
2. **Atomic Number Calculation:**
\[ 92 = 2 + (atomic \, number \, of \, X) \]
Therefore,
\[ atomic \, number \, of \, X = 92 - 2 = 90 \]
Hence, the missing element \(\text{X}\) has an atomic number of 90, which corresponds to Thorium (Th) in the periodic table.
### Completed Reaction:
\[ \phantom{^{238}_{92}}\text{U} \rightarrow \phantom{^{4}_{2}}\text{He} + \phantom{^{234}_{90}}\text{Th} \]
This example shows how the conservation laws for mass and atomic number can help solve for unknowns in nuclear reactions.
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- Consider the fission reaction n + 235 U After having found the missing nuclide, calculate the Q-value of the reaction. Enter your result in MeV to 1 decimal place ⁰Xe + ? + 2n. 140-arrow_forward2H is a loosely bound isotope of hydrogen. Calleddeuterium or heavy hydrogen, it is stable but relativelyrare—it is 0.015% of natural hydrogen. Note that deuteriumhas Z = N , which should tend to make it more tightlybound, but both are odd numbers. Calculate BE/A , thebinding energy per nucleon, for 2H and compare it with theapproximate value obtained from the graph in Figure.arrow_forwardMagnesium (12Mg) has three stable isotopes with consecutive mass numbers. Write the complete nuclear symbol (with neutron number) for each. (Enter your answer in the form AZX_{N}. Separate substances in a list with a comma.)arrow_forward
- #11.arrow_forward(a) In addition to uranium, plutonium is also a well-suited starting product for the release of energy through nuclear fission. If you bombard a ²32Pu nucleus with a neutron the fission products 94 barium ¹Ba and strontium 3Sr are formed. 56 (i) Write down the full equation of the reaction. (ii) Calculate the reaction energy in eV units. Note that m(232Pu) = 239.052162u, 94 m(¹Ba) = 143.922955u, and m(Sr) = 93.915356u. (iii) Explain what triggers a chain reaction with 232 Pu. 94arrow_forward239 239Pu can be bombarded by a neutron and undergo nuclear fission to make an atom of 204 Au and an atom of 3lp. Write the balanced nuclear equation for the process. Make sure to show how many neutrons are given off by this reaction. Include the FULL isotope notation for ALL species involved.arrow_forward
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