An electron is to be accelerated from a velocity of 1.50×106 m/s to a velocity of 9.00×106 m/s . through what potential difference must the electron pass to accomplish this?

Velocity of Electron V1 = 1.50 Ă—10^6 m/s Velocity of Electron V2 = 9.00 Ă— 10^6 m/s We know the mass of electron m = 9.11 x 10^-31 Charge of the electron q = 1.60 x 10^-19 Calculating the potential energy at the two stages, K1 = 0.5 x m V1^2 = 0.5 x 9.11 x 10^-31 x (1.50 Ă—10^6)^2 = 10.24 x 10^-19 K2 = 0.5 x m V2^2 = 0.5 x 9.11 x 10^-31 x (9.00 Ă—10^6)^2 = 368.95 x 10^-19 Change in Potential energy delta K= K2 - K1 = 358.71 x 10^-19 As we know for total energy conservation total change in energy is 0 so q x delta V + delta K = 0 delta V = - delta K / q = 358.71 x 10^-19 / 1.60 x 10^-19 = 224.19V Potential difference for the electron to pass = 224.19V

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