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An electron moves in the lab with a speed of 0.6c. An observer moves with a velocity of 0.8 c along the direction of the motion of the electron. The kinetic energy of the electron as determined by the observer is _____ MeV (Upto three decimal places)
Correct answer is '0.043'. Can you explain this answer?
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An electron moves in the lab with a speed of 0.6c. An observer moves w...
Using Lorentz velocity transformations, velocity of election in the frame of the observer (S') is
Kinetic energy of the electron as observed by the observer
Kinetic energy of the electron as observed by the observer
Most Upvoted Answer
An electron moves in the lab with a speed of 0.6c. An observer moves w...
Given:
- Speed of the electron in the lab frame = 0.6c
- Velocity of the observer = 0.8c (moving along the direction of the electron's motion)
To find:
- Kinetic energy of the electron as determined by the observer
Concept:
- The relativistic kinetic energy of a particle with mass m and velocity v can be given by the equation:
- K.E. = (γ - 1)mc²
where γ is the Lorentz factor given by γ = 1/√(1 - v²/c²)
Calculation:
- Given that the electron is moving with a speed of 0.6c, we can calculate the Lorentz factor γ as follows:
- γ = 1/√(1 - (0.6c)²/c²)
= 1/√(1 - 0.36)
= 1/√0.64
= 1/0.8
= 1.25
- The kinetic energy of the electron in the lab frame can be calculated using the relativistic kinetic energy equation:
- K.E. = (γ - 1)mc²
where m is the mass of the electron and c is the speed of light.
- Since the mass of the electron does not change, we can ignore it for the calculation. Therefore, the kinetic energy of the electron in the lab frame is:
- K.E. = (1.25 - 1)mc²
= 0.25mc²
- Now, let's calculate the velocity of the electron as observed by the observer:
- The observer is moving with a velocity of 0.8c along the direction of the electron's motion.
- The relative velocity between the observer and the electron can be calculated using the relativistic velocity addition formula:
- v' = (v - u) / (1 - uv/c²)
where v is the velocity of the electron in the lab frame and u is the velocity of the observer.
- Substituting the given values, we get:
- v' = (0.6c - 0.8c) / (1 - (0.6c)(0.8c)/c²)
= (-0.2c) / (1 - 0.48)
= -0.2c / 0.52
= -0.3846c
- The negative sign indicates that the electron is moving in the opposite direction as observed by the observer.
- Now, we can calculate the Lorentz factor γ' for the observer:
- γ' = 1/√(1 - (-0.3846c)²/c²)
= 1/√(1 - 0.1481)
= 1/√0.8519
= 1/0.9239
= 1.082
- Finally, the kinetic energy of the electron as determined by the observer can be calculated using the relativistic kinetic energy equation:
- K.E.' = (γ' - 1)m'c²
where m' is the relativistic mass of the electron observed by the observer.
-
- Speed of the electron in the lab frame = 0.6c
- Velocity of the observer = 0.8c (moving along the direction of the electron's motion)
To find:
- Kinetic energy of the electron as determined by the observer
Concept:
- The relativistic kinetic energy of a particle with mass m and velocity v can be given by the equation:
- K.E. = (γ - 1)mc²
where γ is the Lorentz factor given by γ = 1/√(1 - v²/c²)
Calculation:
- Given that the electron is moving with a speed of 0.6c, we can calculate the Lorentz factor γ as follows:
- γ = 1/√(1 - (0.6c)²/c²)
= 1/√(1 - 0.36)
= 1/√0.64
= 1/0.8
= 1.25
- The kinetic energy of the electron in the lab frame can be calculated using the relativistic kinetic energy equation:
- K.E. = (γ - 1)mc²
where m is the mass of the electron and c is the speed of light.
- Since the mass of the electron does not change, we can ignore it for the calculation. Therefore, the kinetic energy of the electron in the lab frame is:
- K.E. = (1.25 - 1)mc²
= 0.25mc²
- Now, let's calculate the velocity of the electron as observed by the observer:
- The observer is moving with a velocity of 0.8c along the direction of the electron's motion.
- The relative velocity between the observer and the electron can be calculated using the relativistic velocity addition formula:
- v' = (v - u) / (1 - uv/c²)
where v is the velocity of the electron in the lab frame and u is the velocity of the observer.
- Substituting the given values, we get:
- v' = (0.6c - 0.8c) / (1 - (0.6c)(0.8c)/c²)
= (-0.2c) / (1 - 0.48)
= -0.2c / 0.52
= -0.3846c
- The negative sign indicates that the electron is moving in the opposite direction as observed by the observer.
- Now, we can calculate the Lorentz factor γ' for the observer:
- γ' = 1/√(1 - (-0.3846c)²/c²)
= 1/√(1 - 0.1481)
= 1/√0.8519
= 1/0.9239
= 1.082
- Finally, the kinetic energy of the electron as determined by the observer can be calculated using the relativistic kinetic energy equation:
- K.E.' = (γ' - 1)m'c²
where m' is the relativistic mass of the electron observed by the observer.
-
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An electron moves in the lab with a speed of 0.6c. An observer moves with a velocity of 0.8 c along the direction of the motion of the electron. The kinetic energy of the electron as determined by the observer is _____ MeV (Upto three decimal places)Correct answer is '0.043'. Can you explain this answer?
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An electron moves in the lab with a speed of 0.6c. An observer moves with a velocity of 0.8 c along the direction of the motion of the electron. The kinetic energy of the electron as determined by the observer is _____ MeV (Upto three decimal places)Correct answer is '0.043'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An electron moves in the lab with a speed of 0.6c. An observer moves with a velocity of 0.8 c along the direction of the motion of the electron. The kinetic energy of the electron as determined by the observer is _____ MeV (Upto three decimal places)Correct answer is '0.043'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electron moves in the lab with a speed of 0.6c. An observer moves with a velocity of 0.8 c along the direction of the motion of the electron. The kinetic energy of the electron as determined by the observer is _____ MeV (Upto three decimal places)Correct answer is '0.043'. Can you explain this answer?.
An electron moves in the lab with a speed of 0.6c. An observer moves with a velocity of 0.8 c along the direction of the motion of the electron. The kinetic energy of the electron as determined by the observer is _____ MeV (Upto three decimal places)Correct answer is '0.043'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An electron moves in the lab with a speed of 0.6c. An observer moves with a velocity of 0.8 c along the direction of the motion of the electron. The kinetic energy of the electron as determined by the observer is _____ MeV (Upto three decimal places)Correct answer is '0.043'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electron moves in the lab with a speed of 0.6c. An observer moves with a velocity of 0.8 c along the direction of the motion of the electron. The kinetic energy of the electron as determined by the observer is _____ MeV (Upto three decimal places)Correct answer is '0.043'. Can you explain this answer?.
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An electron moves in the lab with a speed of 0.6c. An observer moves with a velocity of 0.8 c along the direction of the motion of the electron. The kinetic energy of the electron as determined by the observer is _____ MeV (Upto three decimal places)Correct answer is '0.043'. Can you explain this answer?, a detailed solution for An electron moves in the lab with a speed of 0.6c. An observer moves with a velocity of 0.8 c along the direction of the motion of the electron. The kinetic energy of the electron as determined by the observer is _____ MeV (Upto three decimal places)Correct answer is '0.043'. Can you explain this answer? has been provided alongside types of An electron moves in the lab with a speed of 0.6c. An observer moves with a velocity of 0.8 c along the direction of the motion of the electron. The kinetic energy of the electron as determined by the observer is _____ MeV (Upto three decimal places)Correct answer is '0.043'. Can you explain this answer? theory, EduRev gives you an
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