Dual Nature of Light

The dual nature of light describes that light has dual nature. It behaves as both particle i.e. corpuscles nature (Energy particle of Planck) and waves nature i.e. electromagnetic waves. In phenomena like the Compton effect, the Photoelectric effect light behaves as the particle, and in the phenomenon of light diffraction, interference, and polarization light behaves as the wave because they are explained based on wave theory. In the microscopic description, if the light is propagating in the medium then the wave nature of light is considered and if light mutually interacts with the matter then the particle nature of the light is considered.

Basic Terminology

• Photon: According to Planck’s quantum theory of radiation, an electromagnetic wave travels in the form of discrete packets called quanta. This one quantum of light radiation is called a photon.
• Photoelectric Effect: The phenomenon of emission of electrons from a metal surface, when electromagnetic radiations of sufficiently high frequency are incident on it, is called the photoelectric effect.
• Compton Effect: The scattering of a photon by an electron is called the Compton effect.
• De-Broglie Waves: The wave associated with the moving particles are called matter waves or De-Broglie waves.

De-Broglie’s Hypothesis

According to De-Broglie as there is symmetry in nature, particles must behave as waves.

For light radiation of wavelength λ, according to the Planck equation,

E = hc/λ, where E is energy, c is the speed of light, λ is wavelength and h is Planck’s constant, 6.63×10^-34 Js.

By Einstein’s mass energy relation

E = mc², where E is energy, m is mass and c is the speed of light. 

From equations (1) and (2), we have         

=> mc² = (hc/λ)

=> mc = (h/λ)

=> λ = h/mc

According to De-Broglie 

If a particle of mass ‘m’ is moving with velocity ‘v’, then the wavelength associated with it is given by,

 λ = h/mv (as p = mv)

 λ = h/p 

where, p is the momentum

The Kinetic Energy of the particle is given by

E_k = p²/(2m)                     [p =√(2mE_k )]

then,  λ in terms of kinetic energy E_k is given by,

 λ = h/√(2mE_k )

If the charged particle q is accelerated by potential difference V, then the wavelength is given by,

E_k = qV

λ = h/√(2mqV)

If the electron e is accelerated by potential difference V, then the wavelength is given by,

λ = h/√(2meV )

Substituting m = 9.1 × 10^-31 kg , h = 6.63 × 10^-34 J-s ,  e = 1.6 × 10^-19 C 

λ = √(150/V) 10^-10

λ = √(150/V) A°

λ = 12.27/√V A°

Quantum Nature of Light

Some phenomena like the photoelectric effect and Compton effect could not be explained by the Wave theory of light. The quantum theory of light is proposed by Einstein who extended Planck’s hypothesis to explain Black Body Radiation. According to the quantum theory of light, the energy of an electromagnetic wave is not distributed continuously over the wavefront. But Planck proposed that an electromagnetic wave travels in the form of discrete packets of energy called Quanta. According to Planck, “light is propagated in the bundles of small energy, each bundle is called photon which possesses energy”. This energy is given by,

E = hν = hc/λ

where,

• ν is frequency
• λ is wavelength of light
• h is Planck’s constant
• c is speed of light

Solved Examples on Dual Nature of Light

Example 1: What is the De-Broglie wavelength associated with an electron, accelerated through a potential difference of 100 Volts?

Answer:

De-Broglie wavelength associated with an electron accelerated through a potential difference of V,

λ = 12.27/√V A°

In the question V = 100 Volts

λ = 12.27/√100 A°

λ = 12.27/10 A°

λ = 1.227 A°

Example 2: Find the De-Broglie wavelength associated with an electron moving with a velocity 0.5c and rest mass = 9.1×10^-31 kg.

Answer: 

De-Broglie wavelength associated with an electron moving with a velocity v

λ = h /mv

In the given question, v = 0.5c = 0.5 × 3×10⁸ = 1.5×10⁸ m/s

λ =  6.6×10^-34/ (9.1×10^-31×1.5×10⁸)

λ = 4.8 × 10^-12 m

Example 3: What will be the De-Broglie wavelength of an electron of energy 400 eV? Given h = 6.6×10^-34 J-s , e = 1.6×10^-19C and m = 9.1×10^-31 kg.

Answer: 

De-Broglie wavelength λ in terms of energy E is given by,

λ = h/√(2mE)

λ = 6.6×10^-34/√(2×9.1×10^-31×400×1.6×10^-19)

λ = 0.611 A°

Example 4. For what kinetic energy of a proton, will the associated De-Broglie wavelength be 16.5 nm? Mass of proton = 1.675×10^-27, h = 6.63×10^-34 J-s.

Answer: 

De-Broglie wavelength in terms of kinetic energy is given by,

λ = h/√(2mE_k )

E_k = h²/(λ²2m)

E_k = ( 6.63×10^-34)²/[(16.5×10^-9)²×2×1.675×10^-27]

E_k = 4.82×10^-25 J

FAQs on Dual Nature of Light

Question 1: State the De-Broglie wavelength formula in terms of momentum.

Answer:  

De-Broglie wavelength in terms of momentum is given by, λ = h/p.

Question 2: What are De-Broglie waves?

Answer:  

The waves associated with the moving particles is called De-Broglie waves.

Question 3: What are photons?

Answer:

According to Planck’s quantum theory of radiation, an electromagnetic wave travels in the form of discrete packets called quanta. This one quantum of light radiation is called a photon.