DE BROGLIE WAVELENGTH

de Broglie wave

  • According to de Broglie the matter like radiation has dual nature and particle like and wave like properties. All the matter particles like electrons, protons, neutrons. Have an associated wave with them called as matter wave or pilot wave or de Broglie wave.                        

The de Broglie Equation

  • In 1924 de Broglie suggest that the particles in motion should exhibit the characteristics of waves. de Broglie further suggested that certain basic physical formula should apply both to waves and particles.
  • If we consider a photon as a wave of frequency v, then. Energy of the photon = E = hv —-(1)
  • If we consider the photon as a particle of mass m, then. Energy of the photon = E = mc² —(2)
  • Equation (2) is according to the special theory of relativity. Here c is the velocity of electromagnetic radiation in vacuum. Equating the two expressions for the energy.  E = hv =  mc² —(3)
  •  As the photon travels in free space with a velocity c, its momentum is given by  p = mc —(4)
  • By substituting m in (4) from (3), we get p=mc=\frac{hv}{c^2}c=\frac{hv}{c}=\frac{h}{\lamb} ----(5)   where v=c/λ
  • Where λ is the wavelength of radiation. de Broglie suggested that this equation applied as well to particle of mass m moving with a velocity v. Thus for a particle the momentum is given by p=mv=\frac{h}{\lamb}    ---(6)
  •  Where λ is the wavelength of the waves associated with a particle of mass m moving with velocity v. Hence this wave is known as matter waves. Does the wavelength of this matter wave is given by\lamb=\frac{h}{mv} ---(7). This equation is known as the de broglie equation.

de Broglie Wavelength in terms of kinetic energy (K.E)

The kinetic energy of a moving particle of mass m and velocity v is given by   E=\frac{1}{2}mv^2=\frac{1}{2m}m^2v^2

 E=\frac{p^2}{2m}  , where p=mv. Then p=\sqrt{2mE}  and substituting p value ofequation (6), we get                                                                \lamb=\frac{h}{\sqrt{2mE}}---(8)                   

 

dE Broglie Wavelength of electron

  • Latest consider an electron of mass m, charge e and is accelerated through a potential of V volts. The energy acquired by the electron is eV and is  equal to \frac{1}{2}mv^2
  • \frac{1}{2}=mv^2=eV\Rightarrow {mv^2}=2eV\Rightarrow{m^2v^2}=2meV
  • Then mv=\sqrt{2meV} . Substituting mv value in equation (7), we get \lamb=\frac{h}{\sqrt{2meV}} ---(9)                                                                    
  • knowing h, m and e, λ  becomes  \lamb=\frac{h}{\sqrt{2meV}}=\frac{6.625\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times1.602\times10^{-19}\times V
  • \lamb=\frac{12.24\times10^{-10}}{\sqrt{V}}m ---(10)
  • Equation(10)is the de Broglie wave length of electrons

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