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The wavelength lambda of a photon and th...

The wavelength `lambda` of a photon and the de-Broglie wavelength of an electron have the same value. Show that the energy of the photon is `(2 lambda mc)/h` times the kinetic energy of the electron, Where m,c and h have their usual meanings.

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Energy of a photon , `E="hc"/lambda`
Energy of electron `E_k=1/2mv^2 =1/2 (m^2v^2)/m =1/2 h^2/(mlambda_e^2) [ because lambda=h/"mv"]`
`"hc"/lambda_p=((2lambda_p mc)/h)xx1/2h^2/(mlambda_e^2)`
`lambda_p=lambda_e`
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