The wavelength `lambda` of a photon and the de-Broglie wavelength of an
Channel: Doubtnut
Date: 2020-08-02

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.