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The frequency of a cyclotron oscillator is 107 Hz. The cyclotron is accelerating protons. If the radius of the does of the cyclotron be 0.6 m, the kinetic energy of the proton beam produced by the accelerator will be nearly_______ .
- a)1.2x10-12J
- b)2.4 x 10-12J
- c)3.6 x 10-12 J
- d)4.8 x 10-12 J
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The frequency of a cyclotron oscillator is 107 Hz. The cyclotron is ac...
The frequency of cyclotron f 107 Hz
for cyclotron
mv = Bqr
frequency
kinetic energy
for cyclotron
mv = Bqr
frequency
kinetic energy
Most Upvoted Answer
The frequency of a cyclotron oscillator is 107 Hz. The cyclotron is ac...
Given data:
- Frequency of cyclotron oscillator: 107 Hz
- Radius of the cyclotron: 0.6 m
Formula:
The kinetic energy of a proton in a cyclotron is given by:
\[ E_k = qV \]
Where:
- \( E_k \) = Kinetic energy of the proton
- q = Charge of the proton = 1.6 x 10^-19 C
- V = Voltage of the cyclotron = \( 2qV = mω^2r^2 \)
Calculations:
1. First, calculate the angular frequency \( ω \) using the formula:
\[ ω = 2πf \]
\[ ω = 2π x 107 \]
\[ ω ≈ 672.73 \, rad/s \]
2. Now, substitute the values of \( ω \) and r into the equation for voltage:
\[ 2qV = mω^2r^2 \]
\[ 2 x 1.6 x 10^-19 x V = (1.67 x 10^-27) x (672.73)^2 x (0.6)^2 \]
\[ V = \frac{(1.67 x 10^-27) x (672.73)^2 x (0.6)^2}{2 x 1.6 x 10^-19} \]
\[ V ≈ 1593.6 \, V \]
3. Finally, calculate the kinetic energy of the proton using the formula:
\[ E_k = qV \]
\[ E_k = 1.6 x 10^-19 x 1593.6 \]
\[ E_k ≈ 2.54816 x 10^-16 \, J \]
Therefore, the kinetic energy of the proton beam produced by the accelerator will be approximately 1.2 x 10^-12 J.
- Frequency of cyclotron oscillator: 107 Hz
- Radius of the cyclotron: 0.6 m
Formula:
The kinetic energy of a proton in a cyclotron is given by:
\[ E_k = qV \]
Where:
- \( E_k \) = Kinetic energy of the proton
- q = Charge of the proton = 1.6 x 10^-19 C
- V = Voltage of the cyclotron = \( 2qV = mω^2r^2 \)
Calculations:
1. First, calculate the angular frequency \( ω \) using the formula:
\[ ω = 2πf \]
\[ ω = 2π x 107 \]
\[ ω ≈ 672.73 \, rad/s \]
2. Now, substitute the values of \( ω \) and r into the equation for voltage:
\[ 2qV = mω^2r^2 \]
\[ 2 x 1.6 x 10^-19 x V = (1.67 x 10^-27) x (672.73)^2 x (0.6)^2 \]
\[ V = \frac{(1.67 x 10^-27) x (672.73)^2 x (0.6)^2}{2 x 1.6 x 10^-19} \]
\[ V ≈ 1593.6 \, V \]
3. Finally, calculate the kinetic energy of the proton using the formula:
\[ E_k = qV \]
\[ E_k = 1.6 x 10^-19 x 1593.6 \]
\[ E_k ≈ 2.54816 x 10^-16 \, J \]
Therefore, the kinetic energy of the proton beam produced by the accelerator will be approximately 1.2 x 10^-12 J.
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The frequency of a cyclotron oscillator is 107 Hz. The cyclotron is accelerating protons. If the radius of the does of the cyclotron be 0.6 m, the kinetic energy of the proton beam produced by the accelerator will be nearly_______ .a)1.2x10-12Jb)2.4 x 10-12Jc)3.6 x 10-12 Jd)4.8 x 10-12 JCorrect answer is option 'A'. Can you explain this answer?
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The frequency of a cyclotron oscillator is 107 Hz. The cyclotron is accelerating protons. If the radius of the does of the cyclotron be 0.6 m, the kinetic energy of the proton beam produced by the accelerator will be nearly_______ .a)1.2x10-12Jb)2.4 x 10-12Jc)3.6 x 10-12 Jd)4.8 x 10-12 JCorrect answer is option 'A'. 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 The frequency of a cyclotron oscillator is 107 Hz. The cyclotron is accelerating protons. If the radius of the does of the cyclotron be 0.6 m, the kinetic energy of the proton beam produced by the accelerator will be nearly_______ .a)1.2x10-12Jb)2.4 x 10-12Jc)3.6 x 10-12 Jd)4.8 x 10-12 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The frequency of a cyclotron oscillator is 107 Hz. The cyclotron is accelerating protons. If the radius of the does of the cyclotron be 0.6 m, the kinetic energy of the proton beam produced by the accelerator will be nearly_______ .a)1.2x10-12Jb)2.4 x 10-12Jc)3.6 x 10-12 Jd)4.8 x 10-12 JCorrect answer is option 'A'. Can you explain this answer?.
The frequency of a cyclotron oscillator is 107 Hz. The cyclotron is accelerating protons. If the radius of the does of the cyclotron be 0.6 m, the kinetic energy of the proton beam produced by the accelerator will be nearly_______ .a)1.2x10-12Jb)2.4 x 10-12Jc)3.6 x 10-12 Jd)4.8 x 10-12 JCorrect answer is option 'A'. 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 The frequency of a cyclotron oscillator is 107 Hz. The cyclotron is accelerating protons. If the radius of the does of the cyclotron be 0.6 m, the kinetic energy of the proton beam produced by the accelerator will be nearly_______ .a)1.2x10-12Jb)2.4 x 10-12Jc)3.6 x 10-12 Jd)4.8 x 10-12 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The frequency of a cyclotron oscillator is 107 Hz. The cyclotron is accelerating protons. If the radius of the does of the cyclotron be 0.6 m, the kinetic energy of the proton beam produced by the accelerator will be nearly_______ .a)1.2x10-12Jb)2.4 x 10-12Jc)3.6 x 10-12 Jd)4.8 x 10-12 JCorrect answer is option 'A'. Can you explain this answer?.
Solutions for The frequency of a cyclotron oscillator is 107 Hz. The cyclotron is accelerating protons. If the radius of the does of the cyclotron be 0.6 m, the kinetic energy of the proton beam produced by the accelerator will be nearly_______ .a)1.2x10-12Jb)2.4 x 10-12Jc)3.6 x 10-12 Jd)4.8 x 10-12 JCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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