## Question

A light beam of wavelength 6000 Å and intensity falls normally on a photon-cathode of surface area 1 cm^{2} and work function 2 eV. Assuming that there is no loss of light by reflection etc., calculate the number of photoelectrons emitted per second.

### Solution

10^{12}

The energy of a photon of wavelength is

.

= 2.06 eV.

The work function of the surface is 2 eV. Thus, the energy of photon is greater than the work function of the surface. Hence, the incident light is able to emit photoelectrons.

The number of photons falling on the surface of area 1 cm^{2} (= 10^{–4} m^{2}) of the photocathode is

= 10^{12} s^{–1}.

Assuming that each photon emits one photoelectron, the number of photoelectrons emitted per second is 10^{12}.

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