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 Me V`
B. `2.55 Me V`
C. `20.4 Me V`
D. `5.1 Me V`

Answer

D

Explanation

`T = (1)/(f) = (2pi R)/(v)`
`v = 2 piRf = 2pi xx 0.5 xx 10^(7) = pi xx 10^(7) = pi xx 10^(7) m//sec`
`K = (1)/(2)mv^(2) = (1)/(2) xx 1.67 xx 10^(-27) xx pi^(2) xx 10^(14)`
`= 8.23 xx 10^(-13) J`
`= (8.23 xx 10^(-13))/(1.6 xx 10^(-13))MeV`
`= 5.1 Me V`