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Wave-Matter Duality of Radiation

Wave-matter duality of radiation
In certain experimental observation radiation behaves as wave, i.e it has certain wavelength, frequency and energy.
For example: Experiments associated with
1. Interference
2. Diffraction
3. Polarization etc
In some experimental observation radiation behaves as a stream of particle ( i.e photon ) having dynamic mass, linear momentum and energy.
For example:
1 Compton effect
2. Photoelectric effect
3. Discrete emission and absorption of radiation by black body etc.
The some entity (radiation ) behaves as wave as well as particle. This nature of radiation is known as wave-matter duality of radiation.
According to Plank's theory of radiation energy of photon of frequency $\nu$ is given by:
$$E= h\nu\dotsm(1)$$
Where, h= Planck's Constant ($6.626\times 10^{-34} JS$)
According to Einstein mass energy equivalent relation:
$$E= mc^2\dotsm(2)$$
Where,
m= dynamic mass of photon
c= Speed of photon in free space
From equation (1) and (2)
$$h\nu=mc^2$$
$$or\;\; h\frac{c}{\lambda}= mc^2$$
$$or,\;\; \frac{h}{\lambda}= mc$$
$$\lambda= \frac{h}{mc}= \frac{h}{p}\dotsm(3)$$
Equation (3) connects the wave and particle nature of radiation.
de-Broglie proposed that a moving material particle also has dual nature. The wave associated with moving particle is known as matter- wave or de-Broglie wave.
For a particle of mass m moving with velocity ( Speed ) v is also associated with wave of wavelength $\lambda$.
$$\lambda=\frac{h}{mv}=\frac{h}{p}$$
Where, p is a linear momentum of moving particle.

de-Broglie wavelength of moving particle
A particle which has kinetic energy only is known as free particle. It's potential energy is zero.
$\therefore$ Total energy of free particle is given by
$$E= K.E$$
$$=\frac12 mv^2$$
Where,
m= mass of particle
v= speed of particle
$$\frac{ m^2 v^2}{2m}$$
$$= \frac{p^2}{2m}$$
$$\therefore \;\; p \sqrt{2mE}\dotsm(1)$$
The de-Broglie wavelength of free particle is
$$\lambda = \frac{h}{p}$$
$$=\frac{h}{\sqrt{2mE}}\dotsm(2)$$
$$\Rightarrow \lambda\propto\frac{1}{\sqrt{m}}$$
$$\lambda\propto \frac{1}{\sqrt{E}}$$
the measurement of dde-Broglie wavelength of matter wave is significant only if value of $\lambda$ is comparable to linear dimension of system (particle itself or region where particle is located ).
For example:
Particle of mass (ice) = 1kg
Speed of ice= 10m/s
$$\therefore \lambda= \frac{h}{mv}= \frac{6.62\times 10^{-34}}{1\times 100}$$
$$= 6.62\times 10^{-36} m$$
$v= d^3\;, \frac{m}{v} =\rho$
$m=\rho v$
$1=1000\times d^3$
$d= \biggl(\frac{1}{1000}\biggr)^{\frac13}$
$d= 0.1 m = 10\; cm= 1\times 10^{-1} m$
$\lambda <<< d$ So calculation of $\lambda$ is not important.
Example 2:
de-Broglie wavelenth of electron in 1st orbit of H-atom
Speed of electron $(V_1)=\frac{c}{137}=\frac{3\times 10^8 m/s}{137}$
$$=2.18\times 10^6m/s<<c$$
$$\lambda=\frac{h}{mv}$$
$$=\frac{6.625\times 10^{-34}}{9.1\times 10^{-31}\times 2.18\times 10^6}= 3.15\times 10^{-10}m= 3.15 A^0$$
Radius of first orbit= $0.529\times 10^{-10}m= r$
Circumference =$2\pi\times r= 3.32A^0$
In this case the de-Broglie wave length is comparable to circumference of electric orbit. Hence the measurement of de-Broglie wave length is significant.

de-Broglie wavelength of bounded (non-relativistic particle)

If a particle is moving inside force field it is known as bounded particle. It has both K.E and potential energy.
The total energy of particle is given by
$$E= K.E+ P.E$$
$$or,\;\; E= \frac12 mv^2+ V$$
$$or,\;\; E-V= \frac{p^2}{2m}$$
$$or,\;\; p= \sqrt{2m (E-V)}$$
$\therefore$ de-Broglie wave length ($\lambda)= \frac{h}{p}$
$$\lambda=\frac{h}{\sqrt{2m(E-v)}}$$

Reference

1. Mathews, P.M and K Venkatesan. A Text Book of Quantum Mechanics. New Delhi: Tata McGraw Hill Publishing Co. Ltd, 1997.
2. Merzbacher, E. Quantum Mechanics . New York: John Wiley, 1969.
3. Prakash, S and S Salauja. Quantum Mechanics. Kedar Nath Ram Nath Publishing Co, 2002.
4. Singh, S.P, M.K and K Singh. Quantum Mechanics. Chand & Company Ltd., 2002.