Question

The current of a beam of electrons, each with a speed of 877 m/s, is 7.21 mA. At one point along its path, the beam encounters a potential step of height -1.25 μV. What is the current on the other side of the step boundary?

Answer 1

Ke of the given region,

KE1 = m v^2 / 2 = (9.109 x 10^-31)(877)^2 / 2

= 3.503 x 10^-25 J

angular wave number in this region,

k1 = 2 pi / wavelength

{ wavelength = h / m v}

k1 = 7.575 x 10^6 m^-1

for other region,

KE2 = E - U = (3.503 x 10^-25) - (1.25 x 1.60 x 10^-19 x 10^-6)

= 1.503 x10^-25 J

wave number, k2 = ( 2 pi / h) sqrt(2 m KE2)

k2 = 4.962 x 10^6 m^-1

so reflection coefficient = [1 - (k2/k1)]^2 / [1 + (k2/k1)]^2

= 0.0434

T = 1 - R = 0.957

transmission current = T I

= 7.21 x 0.957

= 6.90 mA ....Ans