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The wavelength of the matter wave can be expressed as
1. \(\frac{h}{{\sqrt {2mE} }}\)
2. \(\frac{h}{{\sqrt {mE} }}\)
3. \(\frac{h}{{\sqrt {3mE} }}\)
4. \(\frac{h}{{mE}}\)

1 Answer

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Correct Answer - Option 1 : \(\frac{h}{{\sqrt {2mE} }}\)

Concept:

de Broglie wavelength of electrons:

  •  Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well,i.e. matter has dual nature.
  • The wavelength of material waves is also known as the de Broglie wavelength.
  • de Broglie wavelength (λ) of electrons can be calculated from Plancks constant h divided by the momentum of the particle
  • So, according to de Broglie, every object has a dual nature- a particle and a wave nature whose wavelength is given by

\(\lambda = \frac{h}{{mv}}\) -- 

or \(\lambda = {h\over p}\) -- (1)

The relationship between momentum and Kinetic Energy is given as

\(p = \sqrt{2mE}\) -- (2)

Calcualtion:

If we put Equation (1) in Equation (2) we get

\(\lambda = {h\over \sqrt{2mE}}\)

So, the correct option is \(\frac{h}{{\sqrt {2mE} }}\)

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