The wavelength lambda , of a photon & the de-broglie wavelength of an electron have the same value.show that the energy of the photon is (2*lambda*m*c)/h times the kinetic energy of the electron, where m,c,h have their usual meanings..

 

Hi,
Let λ be the wavelength of the photon. It is also the de Broglie wavelength of the electron.
We know that the de Broglie wavelength is given as,

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