When a proton is accelerated in a cyclotron device, maximum energy attained by it is 10 MeV. If radius of ‘Dee’ of the cyclotron is doubled and the magnetic field is doubled, then the maximum energy of on α-particle when accelerated in the cyclotron will be

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20 Mev

160 MeV

240 MeV

80 MeV

answer is C.

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Detailed Solution

Maximum kinetic energy attained by a charged particle when accelerated in the cyclotron is given by $K = \frac{Q^{2} B^{2} R^{2}}{2 m}$

$∴ \frac{K_{α}}{K_{p}} = {(\frac{Q_{α}}{Q_{p}})}^{2} . {(\frac{B_{α}}{B_{p}})}^{2} . {(\frac{R_{α}}{R_{p}})}^{2} . {(\frac{m_{p}}{m_{α}})}^{} = {(2)}^{2} . {(1 / 2)}^{2} . {(2)}^{2} . (4) \Rightarrow K_{α} = 10 \times 16 MeV = 160 MeV .$

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