Wave-Matter duality of radiation

In some experiment radiation behave as wave. For example experiment associated with Interference, Diffraction and polarization. At the same time radiation also behave as well as particle. For example experiment associated with compton effect, photoelectric effect, discrete emission and absorption of radiation by black body etc. So de-Broglie concluded that radiation behaves wave as well as particle. The formula which relates them is \(\lambda=\frac{h}{p}\).

Summary

In some experiment radiation behave as wave. For example experiment associated with Interference, Diffraction and polarization. At the same time radiation also behave as well as particle. For example experiment associated with compton effect, photoelectric effect, discrete emission and absorption of radiation by black body etc. So de-Broglie concluded that radiation behaves wave as well as particle. The formula which relates them is \(\lambda=\frac{h}{p}\).

Things to Remember

  • According to Plank's theory of radiation energy of photon of frequency \(\nu\) is given by:

    $$E= h\nu$$

  • According to Einstein mass energy equivalent relation:

    $$E= mc^2$$

  • 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.

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Wave-Matter duality of radiation

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:

(1). For free 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.

Lesson

Introductory Wave Mechanics

Subject

Physics

Grade

Bachelor of Science

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