Electron released from the cathode CRT is accelerated to the anode by accelerating potential of 2kV in time 0.25 microseconds. Taking the mass of the electron m = 9.1 x 10^-31kg and electron charge e = 1.6 x 10^-19C .Determine:
(I) Velocity of the electrons Soln Velocity of electrons V = {√(2eEa)/(m)} = {√ (2 x 1.6 x10^-19 x 2000)/(9.1 x 10-31)} = 26519741.77 m/s (II) Acceleration of the electron Soln Acceleration = change in velocity/time = (26519741.77 - 0)/(0.26x 10-6) = 1.01999 x 10^14 m/s^2 (III) Loss in potential energy Loss in potential energy P.E = eEa = (1.6 x10^-19) x ( 2000) = 3.2 x 10^-16 Joules (III) Gain in Kinetic energy K.E = 1/2(m x V^2) = 1/2( 9.1 x 10^-31 x 26519741.77^2) = 3.2 x 10^-16 Joules