A cyclotron is used to accelerate protons ( ₁¹ H ) deuterons ( ₁² H ) and -particles ( ₂⁴ He ) . While exiting under similar conditions, the minimum KE is gained by

Step-by-step solution

When a charged particle is accelerated by a potential difference V volts, then the radius of the trajectory of the path of the particle is given as, aligned r &= 2 K m B q = 2 q V m B q K &= 1 2 B² q² r² m or K q² m ...(i) aligned For protons ( ₁¹ H ) , deuterons ( ₁² H ) and -particles ( ₂⁴ He ) , aligned q_ p : q_ d : q_ &=1: 1: 2 ...(ii) m_ p : m_ d : m_ &=1: 2: 4 ...(iii) aligned From Eqs. (i), (ii) and (iii), we can write K_ p : K_ d : K_ =1: 1 4 : 1 Thus, minimum kinetic energy is gained by deuterons.