The oscillating frequency of a cyclotron is `10 MHz`. If the radius of its Dees is `0.5 m`, the kinetic energy of a proton which is accelerated by the cyclotron is

Options

A. 10.2 MeV
B. 2.55 MeV
C. 20.4 MeV
D. 5.1 MeV

Answer

D

Explanation

f = 10 MHz , r = 0.5 m
KE = `(1)/(2) mv^(2) = (1)/(2) (B^(2) q^(2) r_(0)^(2))/(m)`
`implies KE = (1)/(2) m ((qB)/(m))^(2) r_(0)^(2) = (1)/(2) m(2nf)^(2) r_(0)^(2) `[ as `f = (Bq)/(2pi m)]`
`implies KE = (1)/(2) xx 1.67 xx 10^(-27) xx 4pi^(2) xx (10^(7))^(2) xx (0.5)^(2)`
`= 8.24 xx 10^(-13)= 5.1` MeV