A cyclotron's oscillator frequency is ( 10 MHz ) and the operating magnetic field 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

Step-by-step solution

We know, radius of circular path r= m v q B v= q B r m (1) Frequency of cyclotron, f= q B 2 m B= 2 m f q (2) Using Eq. (1), kinetic energy = 1 2 m v²= 1 2 m q² B² r² m² Using Eq. (2), kinetic energy = 1 2 q² r² m (2 )² m² f² q² =2 ² f² r² m Therefore, kinetic energy =2 (3.14)² (10 10⁶ )² (60 10⁻² )² (1.67 10⁻²⁷ )=1.13 10⁻¹² ~J Hence, kinetic energy = 1.13 10⁻¹² 1.16 10⁻¹⁹ 10⁶ 7.1 MeV 7 MeV