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A particle leaving a cyclotron has a total relativistic energy of 10 GeV and a relativistic momentum of 8 GeV. What is the rest mass of this particle?
- a)1.20 GeV/c2
- b)0.25 GeV/c2
- c)6 GeV/c2
- d)2.00 GeV/c2
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A particle leaving a cyclotron has a total relativistic energy of 10Ge...
Total energy = 10 GeV
p = 8 GeV
In units in which c = 1 option (d)
p = 8 GeV
In units in which c = 1 option (d)
The correct answer is: 6 GeV/c2
Most Upvoted Answer
A particle leaving a cyclotron has a total relativistic energy of 10Ge...
Given:
Total relativistic energy (E) = 10 GeV
Relativistic momentum (p) = 8 GeV
To Find:
Rest mass of the particle
Formula:
The relativistic energy of a particle is given by the equation:
E = √(p^2c^2 + m^2c^4)
where,
E = relativistic energy
p = relativistic momentum
m = rest mass of the particle
c = speed of light in vacuum (3 x 10^8 m/s)
Let's solve the equation to find the rest mass.
Solution:
We are given the relativistic energy (E) and relativistic momentum (p) of the particle. We can use the formula mentioned above to find the rest mass (m).
E = √(p^2c^2 + m^2c^4)
Substituting the given values:
10 GeV = √((8 GeV)^2c^2 + m^2c^4)
Simplifying the equation:
100 GeV^2 = (64 GeV^2c^2 + m^2c^4)
Rearranging the equation:
m^2c^4 = 100 GeV^2 - 64 GeV^2c^2
m^2c^4 = 100 GeV^2 - 64 GeV^2(3 x 10^8 m/s)^2
m^2c^4 = 100 GeV^2 - 64 GeV^2(9 x 10^16 m^2/s^2)
m^2c^4 = 100 GeV^2 - 576 GeV^2 x 10^16 m^2/s^2
m^2c^4 = (100 - 576 x 10^16) GeV^2 x m^2/s^2
m^2c^4 = (-576 x 10^16) GeV^2 x m^2/s^2
Taking the square root on both sides:
mc^2 = √((-576 x 10^16) GeV^2 x m^2/s^2)
mc^2 = √((-576 x 10^16) GeV^2) x √(m^2/s^2)
mc^2 = √((-576 x 10^16) GeV^2) x m/s
mc^2 = √(-576 x 10^16) GeV x m/s
mc^2 = (24i√10^16) GeV x m/s
As we can see, the square root of a negative number is imaginary (i). This implies that the rest mass of the particle is imaginary, which is not physically possible.
Hence, there seems to be an error in the given information or calculations, as the rest mass cannot be determined with the given values.
Total relativistic energy (E) = 10 GeV
Relativistic momentum (p) = 8 GeV
To Find:
Rest mass of the particle
Formula:
The relativistic energy of a particle is given by the equation:
E = √(p^2c^2 + m^2c^4)
where,
E = relativistic energy
p = relativistic momentum
m = rest mass of the particle
c = speed of light in vacuum (3 x 10^8 m/s)
Let's solve the equation to find the rest mass.
Solution:
We are given the relativistic energy (E) and relativistic momentum (p) of the particle. We can use the formula mentioned above to find the rest mass (m).
E = √(p^2c^2 + m^2c^4)
Substituting the given values:
10 GeV = √((8 GeV)^2c^2 + m^2c^4)
Simplifying the equation:
100 GeV^2 = (64 GeV^2c^2 + m^2c^4)
Rearranging the equation:
m^2c^4 = 100 GeV^2 - 64 GeV^2c^2
m^2c^4 = 100 GeV^2 - 64 GeV^2(3 x 10^8 m/s)^2
m^2c^4 = 100 GeV^2 - 64 GeV^2(9 x 10^16 m^2/s^2)
m^2c^4 = 100 GeV^2 - 576 GeV^2 x 10^16 m^2/s^2
m^2c^4 = (100 - 576 x 10^16) GeV^2 x m^2/s^2
m^2c^4 = (-576 x 10^16) GeV^2 x m^2/s^2
Taking the square root on both sides:
mc^2 = √((-576 x 10^16) GeV^2 x m^2/s^2)
mc^2 = √((-576 x 10^16) GeV^2) x √(m^2/s^2)
mc^2 = √((-576 x 10^16) GeV^2) x m/s
mc^2 = √(-576 x 10^16) GeV x m/s
mc^2 = (24i√10^16) GeV x m/s
As we can see, the square root of a negative number is imaginary (i). This implies that the rest mass of the particle is imaginary, which is not physically possible.
Hence, there seems to be an error in the given information or calculations, as the rest mass cannot be determined with the given values.
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A particle leaving a cyclotron has a total relativistic energy of 10GeVand a relativistic momentum of 8GeV. What is the rest mass of this particle?a)1.20GeV/c2b)0.25GeV/c2c)6GeV/c2d)2.00GeV/c2Correct answer is option 'C'. Can you explain this answer?
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A particle leaving a cyclotron has a total relativistic energy of 10GeVand a relativistic momentum of 8GeV. What is the rest mass of this particle?a)1.20GeV/c2b)0.25GeV/c2c)6GeV/c2d)2.00GeV/c2Correct answer is option 'C'. 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 A particle leaving a cyclotron has a total relativistic energy of 10GeVand a relativistic momentum of 8GeV. What is the rest mass of this particle?a)1.20GeV/c2b)0.25GeV/c2c)6GeV/c2d)2.00GeV/c2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle leaving a cyclotron has a total relativistic energy of 10GeVand a relativistic momentum of 8GeV. What is the rest mass of this particle?a)1.20GeV/c2b)0.25GeV/c2c)6GeV/c2d)2.00GeV/c2Correct answer is option 'C'. Can you explain this answer?.
A particle leaving a cyclotron has a total relativistic energy of 10GeVand a relativistic momentum of 8GeV. What is the rest mass of this particle?a)1.20GeV/c2b)0.25GeV/c2c)6GeV/c2d)2.00GeV/c2Correct answer is option 'C'. 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 A particle leaving a cyclotron has a total relativistic energy of 10GeVand a relativistic momentum of 8GeV. What is the rest mass of this particle?a)1.20GeV/c2b)0.25GeV/c2c)6GeV/c2d)2.00GeV/c2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle leaving a cyclotron has a total relativistic energy of 10GeVand a relativistic momentum of 8GeV. What is the rest mass of this particle?a)1.20GeV/c2b)0.25GeV/c2c)6GeV/c2d)2.00GeV/c2Correct answer is option 'C'. Can you explain this answer?.
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