 # Dual Behavior of Electromagnetic Radiation

## Dual Behavior of Electromagnetic Radiation:

Orginally Electromagnetic Radiation was supposed to be having wave nature only and we could explain phenonmenon like interference and diffraction with the help the wave nature.
But Wave nature of Electromagnetic Radiation could not explain certain things like Black body radiation & photoelectric effect.

In 1900 , Planck Proposed the quantum theory and was able to satisfactory explain the blackbody radiation.As per the theory atoms or molecules emit or absorb energy only in discrete amounts called quantum and not in a continuous manner. Quantum is the smallest amount of energy that is emitted or absorbed in the form of electromagnetic radiation.
Later on Einstein explain the Photoelectric effect with the help of Planck Quantum theory . He suggested beam of light on a metal surface can be viewed as shooting the metal with a beam of particles or photons

Now With both the Planck Quantum theory & Einstein Theory of Photoelectric effect, it was found that Electromagnetic Radiation behaves like particles(photon) also. Now the particle nature was not consistent with the known wave behaviour of light. This caused significant confusion in the scientific cummunity.

The only way to resolve this dilemma is to accept the dual nature of the Electromagnetic Radiation i,e EM posses both the wave nature and particle nature.

Important points to Note
(a) light behaves either as a wave or as a stream of particles depending on the experiment
(b) Whenever radiation interacts with matter, it displays particle like properties in contrast to the wavelike properties (interference and diffraction), which it exhibits when it propagates.

Question 1
A proton of wavelength 400 nm strikes metal surface. The electrons are ejected with velocity 5.85 × 105 m/s. Calculate min. energy required to remove electron from metal surface. (Mass of electron = 9.1 × 10-31 kg)
Solution
Given:-
λ = 400 nm
Velocity = 5.85 × 105 m/s
C = νλ

3 × 108 m/s = ν. 400 × 10(-9) m
or
ν= 7.5 × 10(14) m
Now E = hν
= 6 × 10-34 × 7.5 × 1014
= 45 ×10-20 J
Kinetic Energy of Electron =1/2 mv2
=.5 × 9.1 × 10-31 × 5.85 × 105 × 5.85 × 105
=15.57 ×10-20 J
Min Energy = 45 ×10-20 J - 15.57 ×10-20 J = 29.43 ×10-20 J

Question 2
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol-1.
Solution
From E= h ν or hc/ λ, we have λ = 242nm or 242 x 10-9m
C = 3 × 108 m/s
h = 6.62 × 10-34 Js
Now putting these values in the equation we get
E= (6.62 × 10-34 Js × 3 × 108 m/s) / (242 × -9m)
= 0.0821 × 10-17 J/atom
Or
E = (0.0821 x 10-17 ) / (1000 x 6.02 x 1023)
= 494 KJ/mol
Question 3
Arrange the following types of radiation in increasing order of their wavelength
(a) From microwave oven
(b) AMBER light from traffic signal
(d) Cosmic rays from outer space
(e)X - rays.
Solution
Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio