The oscillating frequency of a cyclotron is 1 | vedclass

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(D) The kinetic energy $(E_k)$ of a charged particle in a cyclotron is given by the formula: $E_k = \frac{q^2 B^2 r^2}{2m}$.The frequency of the cyclo

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$5.1$

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(D) The kinetic energy $(E_k)$ of a charged particle in a cyclotron is given by the formula: $E_k = \frac{q^2 B^2 r^2}{2m}$.The frequency of the cyclotron is given by: $f = \frac{qB}{2 \pi m}$,which implies $qB = 2 \pi m f$.Substituting $qB$ into the kinetic energy formula:$E_k = \frac{(2 \pi m f)^2 r^2}{2m} = 2 \pi^2 m f^2 r^2$.Given values:$m = 1.67 	imes 10^{-27} \, kg$ (mass of proton)$f = 10 	imes 10^6 \, Hz$$r = 0.5 \, m$Calculating $E_k$ in Joules:$E_k = 2 	imes (3.14)^2 	imes (1.67 	imes 10^{-27}) 	imes (10^7)^2 	imes (0.5)^2$$E_k = 2 	imes 9.8596 	imes 1.67 	imes 10^{-27} 	imes 10^{14} 	imes 0.25$$E_k \approx 8.23 	imes 10^{-13} \, J$.Converting to $MeV$:$E_k (in \, eV) = \frac{8.23 	imes 10^{-13}}{1.6 	imes 10^{-19}} \approx 5.14 	imes 10^6 \, eV = 5.14 \, MeV$.Rounding to the nearest option,we get $5.1 \, MeV$.

The oscillating frequency of a cyclotron is $10 \, MHz$. If the radius of its dee is $0.5 \, m$,the kinetic energy of a proton,which is accelerated by the cyclotron,is ...... $MeV$.

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