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Singly charged ions He^(+)are accelerated in a cyclotron so that theirmaximumorbitalradiysis r = 60 cm. The frequencyof a cyclotron'soscillator is equal to v = 10.0 MHz, theeffectiveaccelearatingvolatageacross the deos is V = 50 kV. Neglectingthe gap between the does, find: (a) the total timeof acceleration of the ion, (b) the appoximatedistancecovered by the ion in theprocess of its acceleration. |
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Answer» Solution :(a) The total time of acceleration is, `t = (1)/(2v).n`, where `n` is the number of passage of theDoes. But, `T = "neV" = (B^(2) e^(2) r^(2))/(2m)` or, `n = (B^(2) e r^(2))/(2 mV)` So, `t = (pi)/(eB//m) xx (B^(2) er^(2))/(2 mV) = (pi B r^(2))/(2V) = (pi^(2) mv r^(2))/(eV) = 30 MU s` (b) The distance covered is, `s = sum v_(n). (1)/(2v)` But, `v_(n) = SQRT((2eV)/(m)) sqrt(n)` So, `s = sqrt((eV)/(2MV^(2))) sum sqrt(n) = sqrt((eV)/(2 mv^(2))) f sqrt(n) dn = sqrt((eV)/(2mv^(2))) (2)/(3) n^(3//2)` But, `n = (B^(2) e^(2) r^(2))/(2 eV m) = (2 pi^(2) m V^(2) r^(2))/(eV)` THUS, `s = (4pi^(3)n v^(2) mr^(2))/(3eV) = 1.24 km` |
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