If the maximum value of accelerating potentia | vedclass

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(C) The kinetic energy gained by a proton after $n$ revolutions in a cyclotron is given by $K = n 	imes (2qV)$,where $V$ is the maximum accelerating

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$543$

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(C) The kinetic energy gained by a proton after $n$ revolutions in a cyclotron is given by $K = n 	imes (2qV)$,where $V$ is the maximum accelerating potential per gap and there are two gaps per revolution.Given:$V = 12 	imes 10^3 \, V$$q = 1.6 	imes 10^{-19} \, C$$m_p = 1.67 	imes 10^{-27} \, kg$$v = \frac{c}{6} = \frac{3 	imes 10^8}{6} = 0.5 	imes 10^8 \, m/s$Equating the kinetic energy to the work done:$n(2qV) = \frac{1}{2} m_p v^2$$n = \frac{m_p v^2}{4qV}$$n = \frac{1.67 	imes 10^{-27} 	imes (0.5 	imes 10^8)^2}{4 	imes 1.6 	imes 10^{-19} 	imes 12 	imes 10^3}$$n = \frac{1.67 	imes 10^{-27} 	imes 0.25 	imes 10^{16}}{76.8 	imes 10^{-16}}$$n = \frac{0.4175 	imes 10^{-11}}{76.8 	imes 10^{-16}} \approx 543.6$Rounding to the nearest integer,the number of revolutions is $543$.

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