Define the term ‘intensity of radiation’ in photon picture of light. Ultraviolet light of wavelength 2270 Å from 100 W mercury source irradiates a photo cell made of a given metal. If the stopping potential is – 1 �3 V, estimate the work function of the metal. How would the photo cell respond to a high intensity () red light of wavelength 6300 Å produced by a laser ?

OR

Set up Einstein’s photoelectric equation using the photon picture of electromagnetic radiation. Explain briefly how this equation accounts for all the observations in the photoelectric effect.

A quanta of light is referred to as a photon. The number of such photons, of a particular frequency, incident on a surface per unit time is called Intensity.

Given, wavelength of light

Therefore, kinetic energy of incident radiation,

The stopping potential is

Let be the work function.

We know from Einstein’s photoelectric equation,

Red light has longer wavelength and hence photons have less energy. The minimum energy required for photoelectric emission should be greater than the work function. Even though the intensity is high, there will be no photoelectric emission from red light.

OR

Einstein’s photoelectric equation is based on the particle nature of light. All materials have a given work function(). If an incident photon has energy greater than the work function, it can remove the electron and give it a kinetic energy(KE).

So a part of the incident energy() is used to remove the electron and the rest of the energy imparts a kinetic energy to the electrons.

The Einstein’s photoelectric equation is,

This equation explains that

(i) The kinetic energy of the electrons is independent of the intensity of radiation and only depends on the energy of the photons.

(ii) The energy of the incident photons should be greater than the work function of the material.