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The 238U nucleus has a binding energy of about 7.6 MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100 MeV. From this, it can be concluded that
- a)Nuclei near A = 120 must be bound by about 8.5 MeV/nucleon.
- b)Nuclei near A = 120 must be bound by about 6.7 MeV/nucleon.
- c)238U has a large neutron excess.
- d)Nuclei near A = 120 have masses greater than half that of 238U
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nu...
The binding energy is the energy that keeps a nucleus together and makes its mass slightly different than its constituent particles.
If its masses were exactly the same as its constituent particles, its B.E would be 0 and it would be unstable.
Now, since the nucleus fissions into 2 equal fragmens.
∴ Each would have a mass number half of that of parent nucleus.
i.e. A = 119 for each daughter nucleus. The difference in the Binding energies of the product and parent nucleus would be the K.E of the products.
If we assume that the daughter nucleus (each with A = 119 or A ~ 120) has a B.E of x per nucleon.
(–7.6 + x) 238 = 200
–7.6 + x = 0.83
x = 7.6 + 0.83
x = 8.43
~ 8.5 MeV/nucleus
The correct answer is: Nuclei near A = 120 must be bound by about 8.5 MeV/nucleon.
Most Upvoted Answer
The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nu...
Explanation:
In order to understand why option 'A' is the correct answer, we need to analyze the information given in the question and relate it to the concept of binding energy and kinetic energy in nuclear fission.
Binding Energy:
The binding energy of a nucleus is the energy required to completely separate all of its nucleons (protons and neutrons) from each other. It represents the strength of the nuclear force that holds the nucleus together. The binding energy per nucleon is the binding energy divided by the number of nucleons in the nucleus.
Kinetic Energy in Nuclear Fission:
In nuclear fission, a heavy nucleus splits into two smaller fragments. During this process, a significant amount of energy is released. This energy is distributed as kinetic energy among the fragments. The kinetic energy of each fragment is directly related to the binding energy of the parent nucleus.
Given Information:
- The238Unucleus has a binding energy of about 7.6 MeV/nucleon.
- When the nucleus fissions into two equal fragments, each fragment has a kinetic energy of just over 100 MeV.
Analysis:
1. The binding energy per nucleon of the238Unucleus is 7.6 MeV/nucleon.
2. The total binding energy of the238Unucleus can be calculated by multiplying the binding energy per nucleon by the number of nucleons in the nucleus. Since the number of nucleons in the238Unucleus is 238, the total binding energy is 7.6 MeV/nucleon * 238 nucleons = 1808.8 MeV.
3. When the nucleus undergoes fission, this binding energy is converted into kinetic energy. Since there are two equal fragments, each fragment receives half of the total binding energy, which is 1808.8 MeV / 2 = 904.4 MeV.
4. The question states that each fragment has a kinetic energy of just over 100 MeV. By saying "just over," we can assume that the kinetic energy is approximately 100 MeV.
5. From the above analysis, we can conclude that the binding energy of the fragments is approximately 100 MeV, which is equivalent to 100 MeV / nucleon.
6. The question asks us to find the binding energy per nucleon for nuclei near A=120. Since the binding energy per nucleon for the238Unucleus is 7.6 MeV/nucleon, we can infer that the binding energy per nucleon for nuclei near A=120 is slightly higher.
7. Option 'A' states that nuclei near A=120 must be bound by about 8.5 MeV/nucleon, which is consistent with our analysis.
Therefore, option 'A' is the correct answer.
In order to understand why option 'A' is the correct answer, we need to analyze the information given in the question and relate it to the concept of binding energy and kinetic energy in nuclear fission.
Binding Energy:
The binding energy of a nucleus is the energy required to completely separate all of its nucleons (protons and neutrons) from each other. It represents the strength of the nuclear force that holds the nucleus together. The binding energy per nucleon is the binding energy divided by the number of nucleons in the nucleus.
Kinetic Energy in Nuclear Fission:
In nuclear fission, a heavy nucleus splits into two smaller fragments. During this process, a significant amount of energy is released. This energy is distributed as kinetic energy among the fragments. The kinetic energy of each fragment is directly related to the binding energy of the parent nucleus.
Given Information:
- The238Unucleus has a binding energy of about 7.6 MeV/nucleon.
- When the nucleus fissions into two equal fragments, each fragment has a kinetic energy of just over 100 MeV.
Analysis:
1. The binding energy per nucleon of the238Unucleus is 7.6 MeV/nucleon.
2. The total binding energy of the238Unucleus can be calculated by multiplying the binding energy per nucleon by the number of nucleons in the nucleus. Since the number of nucleons in the238Unucleus is 238, the total binding energy is 7.6 MeV/nucleon * 238 nucleons = 1808.8 MeV.
3. When the nucleus undergoes fission, this binding energy is converted into kinetic energy. Since there are two equal fragments, each fragment receives half of the total binding energy, which is 1808.8 MeV / 2 = 904.4 MeV.
4. The question states that each fragment has a kinetic energy of just over 100 MeV. By saying "just over," we can assume that the kinetic energy is approximately 100 MeV.
5. From the above analysis, we can conclude that the binding energy of the fragments is approximately 100 MeV, which is equivalent to 100 MeV / nucleon.
6. The question asks us to find the binding energy per nucleon for nuclei near A=120. Since the binding energy per nucleon for the238Unucleus is 7.6 MeV/nucleon, we can infer that the binding energy per nucleon for nuclei near A=120 is slightly higher.
7. Option 'A' states that nuclei near A=120 must be bound by about 8.5 MeV/nucleon, which is consistent with our analysis.
Therefore, option 'A' is the correct answer.
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The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer?
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The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer?.
The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer?.
Solutions for The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics.
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Here you can find the meaning of The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The238Unucleus has a binding energy of about 7.6MeV/nucleon. If the nucleus were to fission into 2 equal fragments, each would have a K.E. of just over 100MeV. From this, it can be concluded thata)Nuclei nearA= 120 must be bound by about 8.5MeV/nucleon.b)Nuclei nearA= 120 must be bound by about 6.7MeV/nucleon.c)238Uhas a large neutron excess.d)Nuclei nearA= 120 have masses greater than half that of238UCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Physics tests.
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