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In 1900, Max Planck presented the results of his famous black body radiation experiments, which showed that light has a dual character, behaving like a particle as well as a wave. He gave the Quantum Theory of Radiation explaining electromagnetic radiation and energy.
Its main features were:
- Radiant energy is emitted or absorbed discontinuously in the form of small packets of energy called 'quanta' (and not continuously as thought earlier). Each such quantum is associated with a definite amount of energy. In the case of light, the quantum of energy is called 'photon'.
- The amount of energy associated with a quantum of radiation is proportional to the frequency ( n) of the radiation.
where 'h' is proportionality constant, universally referred to as Planck's constant. It has a fixed value of:
h = 6.63 x 10 34 joule.sec or h = 3.99 x 10 -13 kJ sec mol-1.
This relation was found to be valid for all types of electromagnetic radiation.
- The total amount of energy emitted or absorbed by a body will be some whole number multiple of the quantum, i.e. E
= nh n, where
n = 1, 2, 3, 4, …..
In other words a body can emit or absorb energy equal to 1h, 2h, 3h etc and not as 1.6h n, 2.6h n, 3.2h n etc.
Thus the energy associated with a quantum of radiation depends inversely on its wavelength (or conversely with its frequency) i.e., higher the wavelength of radiation, lesser the energy associated with its quantum. For example a photon of violet light will have more energy than that of a red light because the former has a lower wavelength. The concept of energy packets of light supports the corpuscular character.
Problem
5. Calculate the energy of a photon of light having frequency of
3.0 x 1015 sec-1 (Planck's constant h = 6.63 x 10-34 J s)Solution
The energy of a photon is given by E = h n
where the frequency of light = 3.0 x 1015 s-1Planck's constant, h = 6.63 x 10-34 J s
E = (6.63 x 10-34 J s) x (3.0 x 1015 s-1)= 6.63 x 3 x 1019 J
= 1.99 x 10-18 J.