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A proton moves on a circular path of radius 6.6×10^-3m in a perpendicular magnetic field of 0.625T. the De broglie's wavelength associated with the proton will be.?
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A proton moves on a circular path of radius 6.6×10^-3m in a perpendicu...
Calculation of De Broglie wavelength of a proton in a circular path in a magnetic field
Given:
Radius of circular path, r = 6.6×10^-3m
Magnetic field, B = 0.625T
Formula used:
Magnetic force on a charged particle, F = qvB
Centripetal force, F = mv^2/r
Equating the two forces, we get, mv^2/r = qvB
De Broglie wavelength, λ = h/p
Momentum, p = mv
Calculation:
Using F = qvB and F = mv^2/r, we get, mv^2/r = qvB
Solving for v, we get, v = qBr/m
Momentum, p = mv = mqBr
De Broglie wavelength, λ = h/p = h/mqBr
Substituting the given values, we get,
λ = (6.626×10^-34)/(1.67×10^-27×1.6×10^-19×0.625×6.6×10^-3)
λ ≈ 2.77×10^-12m
Explanation:
A proton moving in a circular path experiences a magnetic force perpendicular to its velocity due to the magnetic field. This magnetic force acts as the centripetal force on the proton and keeps it moving in a circular path. Using the formula for magnetic force and equating it to the centripetal force, we can calculate the velocity of the proton. Using the formula for momentum and De Broglie wavelength, we can then calculate the wavelength associated with the proton.
Given:
Radius of circular path, r = 6.6×10^-3m
Magnetic field, B = 0.625T
Formula used:
Magnetic force on a charged particle, F = qvB
Centripetal force, F = mv^2/r
Equating the two forces, we get, mv^2/r = qvB
De Broglie wavelength, λ = h/p
Momentum, p = mv
Calculation:
Using F = qvB and F = mv^2/r, we get, mv^2/r = qvB
Solving for v, we get, v = qBr/m
Momentum, p = mv = mqBr
De Broglie wavelength, λ = h/p = h/mqBr
Substituting the given values, we get,
λ = (6.626×10^-34)/(1.67×10^-27×1.6×10^-19×0.625×6.6×10^-3)
λ ≈ 2.77×10^-12m
Explanation:
A proton moving in a circular path experiences a magnetic force perpendicular to its velocity due to the magnetic field. This magnetic force acts as the centripetal force on the proton and keeps it moving in a circular path. Using the formula for magnetic force and equating it to the centripetal force, we can calculate the velocity of the proton. Using the formula for momentum and De Broglie wavelength, we can then calculate the wavelength associated with the proton.
Community Answer
A proton moves on a circular path of radius 6.6×10^-3m in a perpendicu...
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A proton moves on a circular path of radius 6.6×10^-3m in a perpendicular magnetic field of 0.625T. the De broglie's wavelength associated with the proton will be.?
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A proton moves on a circular path of radius 6.6×10^-3m in a perpendicular magnetic field of 0.625T. the De broglie's wavelength associated with the proton will be.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A proton moves on a circular path of radius 6.6×10^-3m in a perpendicular magnetic field of 0.625T. the De broglie's wavelength associated with the proton will be.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A proton moves on a circular path of radius 6.6×10^-3m in a perpendicular magnetic field of 0.625T. the De broglie's wavelength associated with the proton will be.?.
A proton moves on a circular path of radius 6.6×10^-3m in a perpendicular magnetic field of 0.625T. the De broglie's wavelength associated with the proton will be.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A proton moves on a circular path of radius 6.6×10^-3m in a perpendicular magnetic field of 0.625T. the De broglie's wavelength associated with the proton will be.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A proton moves on a circular path of radius 6.6×10^-3m in a perpendicular magnetic field of 0.625T. the De broglie's wavelength associated with the proton will be.?.
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