Question

The current of a beam of electrons, each with a speed of 908 m/s, is 7.15 mA. At one point along its path, the beam encounters a potential step of height -1.28 μV. What is the current on the other side of the step boundary?

Answer 1

Given,

drift velocity, v = 908 m/s

current I = 7.15 mA

We know that,

Current, I = (n*e*A)*v

=> 7.15*10^-3 = (n*e*A)*908

=> (n*e*A) = (7.15*10^-3)/(908)

                    = 7.88*10^-6

Now,

there is a potential step of height -1.28 V

Let the new velocity be v'

let m be the mass of electron

=> change in kinetic energy = work done by the electron

=> 1/2*m*v'² - 1/2*m*v² = e*V

=> 1/2*m*(v'² - v²) = e*V

=> (v'² - (908)²) = (e*V)/(1/2*m) = (1.6*10^-19*-1.28*10^-6)/(0.5*9.11*10^-31)

=> (v'² - (908)²) = -0.45*10⁶

=> v'² - 0.83*10⁶ = -0.45*10⁶

=> v'² = -0.45*10⁶ + 0.83*10⁶ = 0.38*10⁶

=> v^, = = 616 m/s

Now,

New current, I' = (n*e*A)*v'

                         = 7.88*10^-6 *616

                         = 4.85*10^-3 = 4.85 mA