State what is meant by the photoelectric effect.

Light of frequency 8.7×10¹⁴Hz is incident on the surface of a metal in a photocell. The surface area of the metal is 9.0×10^–6m² and the intensity of the light is 1.1×10^–3Wm^–2.

(i) Deduce that the maximum possible photoelectric current in the photocell is 2.7 nA.

(ii) The maximum kinetic energy of photoelectrons released from the metal surface is 1.2 eV. Calculate the value of the work function of the metal.

ejection of electron from metal surface following absorption of em radiation/photon;

(i) energy of one photon = 6.67×10⁻³⁴×8.7×10¹⁴(=5.8×10⁻¹⁹J);

number of electrons released from surface per second \( = \frac{{9.0 \times {{10}^{ - 6}} \times 1.1 \times {{10}^{ - 3}}}}{{5.8 \times {{10}^{ - 19}}}}\);
=1.7×10¹⁰;
current=1.7×10¹⁰×1.6×10⁻¹⁹;
= 2.7nA

(ii) 2.4eV or 3.9×10⁻¹⁹J;