Question

A 20 mW 360 nm laser is incident on a metal anode. The metal in the anode has a work
function of 1.80 eV.
a) What voltage must be applied to completely stop the photo-electrons?
b) In the absence of an applied voltage, what current would be measured? assuming
i) all photons in the laser produce a photo-electron and
ii) all photo-electrons reach the cathode
c) Why are these unrealistic assumptions?
d) What is the deBroglie wavelength of the emitted electrons?

Answer 1

For the external photoelectric effect we have

Photon energy = Work function + Electron kinetic energy

h*F = W + Ek

h*F = W +eU

where U is the stopping potential that need to be applied to completely stop all electrons

U =[(h*c/lambda)-W)/ e = 6.62*10^-34*3*10^8/360*10^-9/1.6*10^-19 - 1.8 =1.648 V

If N/t is number of photons in time unit then

Power = Energy/time = (N/t) h*F = (N/t)*(h*c/lambda)

N/t = P*lambda/(h*c) = 0.02*360*10^-9/6.62*10^-34/3*10^8 =3.62*10^16 photons/sec

If 1 photon = 1 electron emitted then current is

I = Q/t = (N/t)*e =3.62*10^16*1.6*10^-19 =5.8*10^-3 A =5.8 mA

These are unrealistic assumption because of two main reasons:

- efficiency of photoelectric effect is ussualy 5-20%.

- electrons emitted in their way to cathode scatter on wir molecules that remain in tube and do not reach the cathode. The vacuum in the tube need to be less than 10^-9 Torr. (Ultra high vacuum)

De Broglie wavelength is

lambda = h/p = h/sqrt(2*m*Ek) = h/sqrt(2*m*e*U) =6.62*10^-34/sqrt(2*9.1*10^-31*1.6*10^-19*1.648) =9.56*10^-10 m