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:

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 (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

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