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Wave Particle Duality

Explore the concept of wave-particle duality in quantum mechanics, where particles exhibit both wave-like and particle-like properties. Understand its implications
authorImageAbhishek Kumar14 Nov, 2023
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Wave Particle Duality

After the work of Planck, Einstein and Compton it was established that radiation has dual nature. It acts like a wave in some situations and as a particle in some other situations.

In 1924, Louis de Broglie extended this wave particle duality to material particles. Nature has two entities - matter and radiation. If radiation has particle-like properties then material particles (electron, proton, neutron etc.) should possess wave-like properties. A photon of frequency has energy,

E=hν

and momentum,

p = E/c = h ν/ c

In terms of wavelength this can be written as,

p = h/λ

where λ= c/ ν De Broglie proposed that this relation applies to both material particles as well as photons. He postulated that with every material particle a wave is associated, having wavelength

λ = h/p

For a particle of mass m, moving with a speed v, wavelength will be

λ = h/mv

This is called de Broglie’s Hypothesis and is called de Broglie wavelength, the associated wave is called matter wave or de Broglie wave.

By the Bohr postulate for stationary orbits, de Broglie tried to explain the stationary orbit by fitting a standing wave around the circumference of an orbit. If is the wavelength associated with nth orbit and r is the radius, then

2 π r = n λ 1

Bohr postulate is,

mvr = nh/2 π→ 2

Putting the value of r from equation 2 in equation 1

n λ = 2 π (nh/2 π mv)

On solving we get,

λ = h/mv

De Broglie wave for n=5 Bohr orbit

We may write,

p = h/ λ = ħk

where, k = 2 π/ λ

E = ħω

p = ħk

Basic relations of Quantum theory

Expression for De Broglie Wavelength:

For Nonrelativistic case:

  • In terms of Kinetic Energy:
For a non-relativistic particle of mass m and kinetic energy K,

De Broglie wavelength can be expressed as,

  • In terms of Accelerating potential:
If a particle of mass m and charge q is accelerated through a potential difference V,

For an electron,

For Relativistic case:

  • In terms of Kinetic Energy:
For a particle of rest mass m0, the total energy is given by

If K is the kinetic energy of the particle, E = K + m 0 c 2 Therefore,

The de Broglie wavelength is,

  • In terms of Accelerating Potential:
If a particle of charge q is accelerated through a potential difference V,

For an electron,

Where,

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