A beam of light of wavelength 400 nm and power 1.55 mW is directed at the cathode of a photoelectric cell. If only 10% of the incident photons produce photoelectrons, what is the photocurrent?

Energy of incident photons = hc/λ = 1240/400 eV = 3.1 eV = 4.96 × 10⁻¹⁹ J

Hence the number of incident photons per second is:
(1.55 × 10⁻³ J/s) / (4.96 × 10⁻¹⁹ J/photon) = 3.125 × 10¹⁵ photons/s

Since only 10% of the incident photons produce photoelectrons, the number of photoelectrons emitted per second is:
0.1 × 3.125 × 10¹⁵ photons/s = 3.125 × 10¹⁴ electrons/s

Hence the photocurrent = e × number of electrons per second = (1.6 × 10⁻¹⁹ C/electron) × (3.125 × 10¹⁴ electrons/s) = 50 × 10⁻⁶ A = 50 μA