Question

Consider electrons at the surface of a metal whose work function is 2.4 eV. Laser light of wavelength 350 nm and power output of 100 mW is shone on this metal, and all the energy of each photon is absorbed by the electrons at the surface. If we model the de Broglie wave function of the electrons as plane waves, how many electrons per second do we expect to see ejected from the metal’s surface? (5)

Hint: treat the work function like a potential step with an associated transmission coefficient to see how many of the excited electrons actually manage to leave the metal.

Answer 1

Energy of a single photon is given by:

E = hc/λ

E = (6.626 x 10^-34 x 3 x 10⁸) / (350 x 10^-9)

E = 5.679 x 10^-19 J

Given that, Work function of the metal:

W_o = 2.4 eV = 2.4 x (1.602 x 10^-19 J) = 3.8448 x 10^-19 J

Hence, Total kinetic energy of the ejected electrons will be:

K = E - W_o

K = (5.679 x 10^-19 J) - (3.8448 x 10^-19 J) = 1.835 x 10^-19 J

Therefore, No.of electrons released per second will be:

N = (100 mJ/s) / (1.835 x 10^-19 J)

N = 5.451 x 10¹⁷ electrons per second

• No.of electrons released per second = 5.451 x 10¹⁷ electrons per second