The wavelength lambda , of a photon the de-broglie wavelength of an electron have the same value show - Physics - Dual Nature Of Radiation And Matter

Question

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

Somnath physics · Accepted Answer

Hi,
Let Î» 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,

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..

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 (2lambdam*c)/h times the kinetic energy of the electron, where m,c,h have their usual meanings..