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प्रश्न
Describe the motion of a charged particle in a cyclotron if the frequency of the radio frequency (rf) field were doubled.
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उत्तर
The frequency va of the applied voltage (radio frequency) is adjusted so that the polarity of the dees is reversed at the same time that it takes the ions to complete one-half of the revolution. The requirement va = vc is called the resonance condition.
When the frequency of the radio frequency (rf) field was doubled, then the resonance condition are violated and the time period of the radio frequency (rf) field was halved. Therefore, the duration in which a particle completes half revolution inside the dees, radio frequency completes the cycle.
So, particles will accelerate and decelerate alternatively. So, the radius of path in the dees will remain the same.
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संबंधित प्रश्न
Which one of the following particles cannot be accelerated by a cyclotron?
(A) Electrons
(B) Protons
(C) Deuterons
(D) α- particles
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Obtain the expression for the cyclotron frequency.
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Give reason to justify your answer.
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Verify that the units weber and volt second are the same.
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Reason: Greater the mass of the particle less will be the frequency of the particle.
A particle of mass m is moving in a circular path of constant radius r such that, its centripetal acceleration ac is varying with time t as ac = k2rt2, where k is a constant. The power delivered to the particle by the forces acting on it is ______.
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Verify that the cyclotron frequency ω = eB/m has the correct dimensions of [T]–1.
