Question

A metal cathode whose work function is 3.3 eV is illuminated with 17 μW of light having a wavelength of 300 nm . The efficiency of converting photons to photoelectrons is 11%. What current is measured in the experiment?

Answer 1

here ,

energy of each photon , E = h * c/(wavelength)

E = 6.626 *10^-34 * 3 *10^8/(300 *10^-9)

E = 6.626 *10^-19 J

NOw, for the number of photons each second

N = total energy/energy of each photon

N = 17 *10^-6/(6.626 *10^-19)

N = 2.57 *10^13

Now , for the current

current = number of electrons * e

current = 2.57 *10^13 * 0.11 * 1.602 *10^-19

current = 4.52 *10^-7 A

the current measured in this experiment is 0.452 μA