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It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :
- a)(.89, .28)
- b)(.28, .89)
- c)(0, 0)
- d)(0, 1)
Correct answer is option 'A'. Can you explain this answer?
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It is found that if a neutron suffers an elastic collinear collision w...
Given information:
- Neutron suffers an elastic collinear collision with deuterium at rest.
- Neutron suffers an elastic collinear collision with carbon nucleus at rest.
To find:
The fractional loss of energy in each collision, i.e., pd and pc.
Solution:
Elastic collision:
An elastic collision is a collision in which both momentum and kinetic energy are conserved.
Collinear collision:
A collinear collision is a collision in which the particles move along the same line before and after the collision.
Collision with deuterium:
Deuterium is the nucleus of a deuterium atom, which consists of a proton and a neutron.
Since the neutron collides with deuterium at rest, the deuterium nucleus does not have any initial kinetic energy.
After the collision, the neutron will lose some of its energy. Let's assume the fractional loss of energy is pd.
Since the collision is elastic and collinear, both momentum and kinetic energy will be conserved.
Let m be the mass of the neutron and v be its initial velocity.
Let M be the mass of the deuterium nucleus.
The initial momentum of the system is mv, and after the collision, the final momentum of the system is (m - pdm)v', where v' is the final velocity of the neutron after the collision.
The initial kinetic energy of the system is (1/2)mv^2, and after the collision, the final kinetic energy of the system is (1/2)(m - pdm)v'^2.
Since the collision is elastic and collinear, momentum conservation gives:
mv = (m - pdm)v'
Simplifying, we get:
v' = v/(1 - pd)
Kinetic energy conservation gives:
(1/2)mv^2 = (1/2)(m - pdm)v'^2
Substituting the value of v', we get:
(1/2)mv^2 = (1/2)(m - pdm)(v/(1 - pd))^2
Simplifying, we get:
pd = ((2m - M)v^2)/(2Mv^2)
Collision with carbon nucleus:
Similarly, let pc be the fractional loss of energy when the neutron collides with a carbon nucleus at rest.
Using the same approach as above, we can derive the equation:
pc = ((2m - M)v^2)/(2Mv^2)
Comparison:
Comparing the expressions for pd and pc, we can see that they are the same.
Therefore, the values of pd and pc are equal.
The correct answer is option A, which is (0.89, 0.28).
- Neutron suffers an elastic collinear collision with deuterium at rest.
- Neutron suffers an elastic collinear collision with carbon nucleus at rest.
To find:
The fractional loss of energy in each collision, i.e., pd and pc.
Solution:
Elastic collision:
An elastic collision is a collision in which both momentum and kinetic energy are conserved.
Collinear collision:
A collinear collision is a collision in which the particles move along the same line before and after the collision.
Collision with deuterium:
Deuterium is the nucleus of a deuterium atom, which consists of a proton and a neutron.
Since the neutron collides with deuterium at rest, the deuterium nucleus does not have any initial kinetic energy.
After the collision, the neutron will lose some of its energy. Let's assume the fractional loss of energy is pd.
Since the collision is elastic and collinear, both momentum and kinetic energy will be conserved.
Let m be the mass of the neutron and v be its initial velocity.
Let M be the mass of the deuterium nucleus.
The initial momentum of the system is mv, and after the collision, the final momentum of the system is (m - pdm)v', where v' is the final velocity of the neutron after the collision.
The initial kinetic energy of the system is (1/2)mv^2, and after the collision, the final kinetic energy of the system is (1/2)(m - pdm)v'^2.
Since the collision is elastic and collinear, momentum conservation gives:
mv = (m - pdm)v'
Simplifying, we get:
v' = v/(1 - pd)
Kinetic energy conservation gives:
(1/2)mv^2 = (1/2)(m - pdm)v'^2
Substituting the value of v', we get:
(1/2)mv^2 = (1/2)(m - pdm)(v/(1 - pd))^2
Simplifying, we get:
pd = ((2m - M)v^2)/(2Mv^2)
Collision with carbon nucleus:
Similarly, let pc be the fractional loss of energy when the neutron collides with a carbon nucleus at rest.
Using the same approach as above, we can derive the equation:
pc = ((2m - M)v^2)/(2Mv^2)
Comparison:
Comparing the expressions for pd and pc, we can see that they are the same.
Therefore, the values of pd and pc are equal.
The correct answer is option A, which is (0.89, 0.28).
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It is found that if a neutron suffers an elastic collinear collision w...
(from conservation of momentum)and K0 = K1 + K2 (for elastic collision)
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It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :a)(.89, .28)b)(.28, .89)c)(0, 0)d)(0, 1)Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :a)(.89, .28)b)(.28, .89)c)(0, 0)d)(0, 1)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :a)(.89, .28)b)(.28, .89)c)(0, 0)d)(0, 1)Correct answer is option 'A'. Can you explain this answer?.
It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :a)(.89, .28)b)(.28, .89)c)(0, 0)d)(0, 1)Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :a)(.89, .28)b)(.28, .89)c)(0, 0)d)(0, 1)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :a)(.89, .28)b)(.28, .89)c)(0, 0)d)(0, 1)Correct answer is option 'A'. Can you explain this answer?.
Solutions for It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :a)(.89, .28)b)(.28, .89)c)(0, 0)d)(0, 1)Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :a)(.89, .28)b)(.28, .89)c)(0, 0)d)(0, 1)Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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