de Broglie hypothesis | Quantum mechanics | Physical basis of quantum mechanics

de Broglie hypothesis

• According to de-Broglie, a wave is always associated with every moving particle. This wave is known as de-Broglie wave or matter wave.
• It means a matter wave have particle nature as well as wave nature i.e., dual nature.
• From quantum theory of radiation, energy of photon, E = h𝝂,   where 𝝂 = frequency of incident photon .
• From Einstein’s theory of relativity E = √(m₀²c⁴ + p²c²)
• If m₀ = 0, then E = pc,  p = momentum and c = velocity of light
• Now E = h𝝂 and E = pc
• ∴     h𝝂 = pc     or     p = h𝝂/c
• ∵      c = 𝝂λ      or     λ = c/𝝂
• ∴      p = h/λ     or     λ = h/p     This is known as de-Broglie wavelength
• This wavelength is always associated with a photon
• Since momentum is the characteristic of particles, and the wavelength is characteristic of wave.
• Hence a wave is always associated with a particle.

Conclusion

• The wavelength of a particle is independent of the charge or the nature of the particle.
• The E.M. waves are produced only due to the charged particle. Hence the matter waves are not E.M. in nature.
• ∵     λ = h/p    and   p = mv      ⇒     λ ∝ 1/v    and   λ ∝ 1/m,      v = velocity of particle 
• Faster is the particle, smaller is its wavelength.
• Heavier is the particle, smaller is its wavelength.
• If v is comparable to c, then m = m₀/√(1 − v²/c²) 

                

de Broglie wavelength of different particles

• If an electron is accelerated by applying a potential difference V, then 
• Energy acquired with it is E = eV
• If m₀ is rest mass of electron, and v is the velocity of electron, then 
• Kinetic energy of electron, E = ½ m₀v²    or   v = √(2E/m₀)
• If relativistic variation of mass with velocity is ignored m ≈ m₀ 
• ∴     v = √(2E/m)
• But  E = eV
• ∴     v = √(2eV/m)
• ∵     de-Broglie wavelength λ = h/mv and v = √(2eV/m)
• ∴     λ = h/√2meV      or      λ = 12.27/√V  Å
• For any charged particle    λ = h/√2mqV

• For any massive particle    λ = h/√2mE            where E is kinetic energy

• If a matter particle at absolute temperature T is in thermal equilibrium, then 
• E = 3/2 kT      where   k = Boltzmann's constant and T is absolute temperature
• ∵     λ = h/√2mE     ∴   λ = h/√3mkT

  Bohr’s hypothesis from de Broglie’s hypothesis

• According to Bohr’s postulate, an electron can revolve around the nucleus only in those orbits whose angular momentum is an integral multiple of h/2π.
• ∴     mvr = nh/2π
• Since a wave is associated with a moving particle, so 2πr = nλ
• Here 2πr is the circumference of the stable orbit
• From de-Broglie hypothesis,  λ = h/mv
• Now 2πr = nλ  and  λ = h/mv
• ∴     2πr = nh/mv ⇒ mvr = nh/2π
• Which is the Bohr's hypothesis

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