A cycolotron's oscillator frequency is 10 MHz and the operating magnetic is 0.66 T . If the radius of its dees is 60 cm, then the kinetic energy of the proton beam produced by the accelerator is

Options

A. 9 MeV
B. 10 MeV
C. 7 MeV
D. 11 MeV

Answer

C

Explanation

Here,
Oscillator frequency, `upsilon=10MHz=10^6Hz`
Radius of the dees , R = 60 cm = 0.6 m
Magnetic field , B = 0.66 T
Charge on the proton , `q=1.6xx10^(-19)C`
Mass of the proton , `m=1.67xx10^(-27)kg`
The kinetic energy of the proton beam produced by the accelerator is
`K=(B^2q^2R^2)/(2m)`
`=((0.66T)^2(1.6xx10^(-19)C)^2(0.6m)^2)/(2(1.67xx10^(-27)kg))`
`=1.2xx10^(-12)J`
`=(1.2xx10^(-12))/(1.6xx10^(-13))MeV(as 1 MeV = 1.6xx10^(-13))`
`=7.5MeV ~~7 MeV`