A cyclotron is used to accelerate protons. If the operating magnetic field is 1 . 0 T and the radius of the cyclotron 'dees' is 60 cm , the kinetic energy of the accelerated protons in MeV will be : [use m p = 1 . 6 × 10 - 27 kg , e = 1 . 6 × 10 - 19 C ]

Step-by-step solution

When the proton is rotated in a cyclotron its maximum velocity will be corresponding to the maximum radius of the circular path. We know that the radius of the circle in a magnetic field is given by, r = m v B q ⇒ v = B q r m . The kinetic energy will be, K = 1 2 m v 2 = B 2 q 2 r 2 2 m ⇒ K = 1 2 × 1 . 6 × 10 - 19 2 × 0 . 6 2 2 × 1 . 6 × 10 - 27 × 1 . 6 × 10 - 13   MeV ⇒ K = 18   MeV