# Particle Nature of Light & Photoelectric Effect

It was observed by Hertz in 1887, quite by accident, that a spark would jump easily between two charged spheres when their surfaces are illuminated by light from another spark.

The photoelectric effect, which is the emission of electricity from metals due to incident electromagnetic radiation, was first investigated in detail by Hallwachs & Lenard during 1886-1900.

The explanation of these experimental results came only after Max Planck proposed the quantum theory of radiation.

It was Sir Isaac Newton who had initially proposed the corpuscular theory of light. His theory was abandoned in favour of the wave theory, proposed by Huygens, as the latter was in agreement with experiments like interference and diffraction.

More than a century later, Planck’s quantum theory (somewhat similar to Newton’s corpuscular theory), got support from Einstein in the explanation of the photoelectric effect .

According to Planck’s quantum theory, light consists of packets of energy, referred to as photons hereafter, which have the following properties:

(i) A photon of light of frequency ν contains energy E which is directly proportional to the frequency :

E = hν  , where h is Planck’s constant

(ii) Photons also carry momentum p :

$\large p = \frac{E}{c} = \frac{h \nu}{c} = \frac{h}{\lambda}$ ;

Where E is the energy of the photon, and c is the velocity of light in vacuum.

(iii) A photon has zero rest mass and moves with the velocity of light in vacuum (c = 3 × 108m/s). It can never be brought to rest.

Energy of Photon:    $\large E = h \nu = \frac{h c}{\lambda}$  ;

In Electron volt , $\large E(eV) = \frac{1242}{\lambda (nm)}$ ; here λ is in nanometer & Energy is in eV .

Mass of Photon : Actually rest mass of photon is zero , but its effective mass is given by :

$\large E = m c^2 = h \nu$

$\large m = \frac{E}{c^2} = \frac{h \nu}{c^2}$  ; This mass is also known as kinetic mass of photon.

Momentum of Photon: $\large p = mc = \frac{E}{c} = \frac{h \nu}{c}= \frac{h}{\lambda}$

Number of photon emitted  per second : $\large n = \frac{P}{E} = \frac{P}{h \nu} = \frac{P \lambda }{h c}$ ; Where E= energy of each photon & P = Power of Light source

Intensity of light: Energy crossing per unit area normally per second is called Intensity or Energy flux .

$\large I = \frac{E}{At} = \frac{P}{A}$ ; Where P = radiation Power ;

At a distance r from point source of power P intensity is given by

$\large I = \frac{P}{4 \pi r^2 }$

When light is incident on a surface , it exerts Pressure

If light of intensity I or power P is incident normally on a surface of area A .

Momentum of incident photons $\large p = \frac{h}{\lambda}$

Change in momentum due to reflection $\large 2 p = \frac{2 h}{\lambda}$

The total momentum imparted to the surface per second is Force .

$\large F = \frac{2 h}{\lambda} \times n$

$\large F = \frac{2 h}{\lambda} \times \frac{P \lambda }{h c} = \frac{2 P}{c}$ ; here P = Power

(a) Pressure , $p = \frac{2I}{c}$ , if light is completely reflected

Pressure $p = \frac{I}{c}$  , if light is completely absorbed .  c = speed of light

(b) $\large F = \frac{2P}{c} = \frac{2IA}{c}$  ; if light is completely reflected

$\large F = \frac{P}{c} = \frac{IA}{c}$  ; if light is completely absorbed

### Einstein explained the photoelectric effect by applying Max Planck’s quantum theory:

(i) Metals contain ” free electrons” which are free to move anywhere within the body of the metal. These “free electrons” however, cannot escape from the body of the metal and are, therefore, bound in that sense. The binding energy of these electrons with the metal varies from one electron to another – the minimum binding energy of a free electron is known as the work function of the metal (W). It is the minimum amount of energy that is required to extract an electron from a metal.

(ii) A single photon of light can cause the emission of a single electron, and no more. Not all photons, however, end up causing the emission of photoelectrons.

The photoelectric effect is not 100% efficient in converting photons into electrons.

(iii) A photon must posses a minimum amount of energy, equal to the work function of the metal, in order to be able to eject a photoelectron.

This means, according to Planck’s quantum theory, there is a minimum possible frequency below which photoelectric effect does not take place. This frequency is also known as the threshold frequency.

If a photon of frequency ν causes the emission of an electron, with a final velocity v, from a metal having work function W, we have, using conservation of energy,

$\large \frac{1}{2}m v^2 = h \nu – B$ ; B =  the binding energy ;

For the fastest photoelectron that is emitted, this becomes,

$\large \frac{1}{2}m v^2 = h \nu – W$ ;

Where, W = hνo, and νo represents the threshold frequency i.e. the minimum frequency of light that can cause the photoelectric effect.

Example : Violet light (λ = 4000A0) of intensity 4 watt/m2 falls normally on a surface of area 10 cm x 20 cm. Find

(a) the energy received by the surface per second,

(b) the number of photons hitting the surface per second,

(c) If surface is tilted such that plane of the surface makes an angle 30° with light beam, find the number of photons hitting the surface per second.

Solution :(a) Energy received per second per unit area

E = IA cos θ

= 4 × 0.02 J = 0.08 J.

(b) n h (c/λ) = E

⇒ n = (0.08 ×4000 × 10-10)/(6.63 × 10-34 × 3 × 108)

= 1.609 × 1017

(c) n = IAcos60° × λ/hc

= 0.805 × 1017