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The amount of work done to increase the speed of an electron from c/3 to 2c/3 is (c = 3×108 m/s and rest mass of electron is 0.511 MeV)
- a)56.50 keV
- b)143.58 keV
- c)168.20 keV
- d)511.00 keV
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The amount of work done to increase the speed of an electron from c/3 ...
To calculate the work done to increase the speed of an electron, we can use the formula for work:
Work = Change in kinetic energy
The initial kinetic energy of the electron is given by:
K1 = (1/2)mv1^2
where m is the mass of the electron and v1 is the initial speed.
The final kinetic energy of the electron is given by:
K2 = (1/2)mv2^2
where v2 is the final speed.
The change in kinetic energy is then:
ΔK = K2 - K1
To find the work done, we substitute the values given:
ΔK = (1/2)mv2^2 - (1/2)mv1^2
Since the mass of an electron is constant, we can simplify the equation to:
ΔK = (1/2)m(v2^2 - v1^2)
Given that v1 = c/3 and v2 = 2c/3, we can substitute these values:
ΔK = (1/2)m((2c/3)^2 - (c/3)^2)
ΔK = (1/2)m((4c^2/9) - (c^2/9))
ΔK = (1/2)m(3c^2/9)
ΔK = (1/2)m(c^2/3)
Therefore, the amount of work done to increase the speed of an electron from c/3 to 2c/3 is (1/2)m(c^2/3), where c = 3.
Work = Change in kinetic energy
The initial kinetic energy of the electron is given by:
K1 = (1/2)mv1^2
where m is the mass of the electron and v1 is the initial speed.
The final kinetic energy of the electron is given by:
K2 = (1/2)mv2^2
where v2 is the final speed.
The change in kinetic energy is then:
ΔK = K2 - K1
To find the work done, we substitute the values given:
ΔK = (1/2)mv2^2 - (1/2)mv1^2
Since the mass of an electron is constant, we can simplify the equation to:
ΔK = (1/2)m(v2^2 - v1^2)
Given that v1 = c/3 and v2 = 2c/3, we can substitute these values:
ΔK = (1/2)m((2c/3)^2 - (c/3)^2)
ΔK = (1/2)m((4c^2/9) - (c^2/9))
ΔK = (1/2)m(3c^2/9)
ΔK = (1/2)m(c^2/3)
Therefore, the amount of work done to increase the speed of an electron from c/3 to 2c/3 is (1/2)m(c^2/3), where c = 3.
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The amount of work done to increase the speed of an electron from c/3 ...
Work done equals to change in kinetic energy. Therefore, Kf minus Ki equals WD.
and use the relativistic formulas of KE equals TEminus RME
and use the relativistic formulas of KE equals TEminus RME
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The amount of work done to increase the speed of an electron from c/3 to 2c/3 is (c = 3×108 m/s and rest mass of electron is 0.511 MeV)a)56.50 keVb)143.58 keVc)168.20 keVd)511.00 keVCorrect answer is option 'B'. Can you explain this answer? for IIT JAM 2025 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The amount of work done to increase the speed of an electron from c/3 to 2c/3 is (c = 3×108 m/s and rest mass of electron is 0.511 MeV)a)56.50 keVb)143.58 keVc)168.20 keVd)511.00 keVCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for IIT JAM 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The amount of work done to increase the speed of an electron from c/3 to 2c/3 is (c = 3×108 m/s and rest mass of electron is 0.511 MeV)a)56.50 keVb)143.58 keVc)168.20 keVd)511.00 keVCorrect answer is option 'B'. Can you explain this answer?.
The amount of work done to increase the speed of an electron from c/3 to 2c/3 is (c = 3×108 m/s and rest mass of electron is 0.511 MeV)a)56.50 keVb)143.58 keVc)168.20 keVd)511.00 keVCorrect answer is option 'B'. Can you explain this answer? for IIT JAM 2025 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The amount of work done to increase the speed of an electron from c/3 to 2c/3 is (c = 3×108 m/s and rest mass of electron is 0.511 MeV)a)56.50 keVb)143.58 keVc)168.20 keVd)511.00 keVCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for IIT JAM 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The amount of work done to increase the speed of an electron from c/3 to 2c/3 is (c = 3×108 m/s and rest mass of electron is 0.511 MeV)a)56.50 keVb)143.58 keVc)168.20 keVd)511.00 keVCorrect answer is option 'B'. Can you explain this answer?.
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