The de Broglie wavelength of an electron and the wavelength of a photon are same. The ratio between the energy of the photon and the momentum of the electron is

The de Broglie wavelength of an electron is the same as that 50 keV X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is 0.5 MeV)

The de-Broglie wavelength of an electron is the same as that of a 50 ke X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is ( the energy equivalent of electron mass of 0.5 MeV)

The de Broglie wavelength of an electron is the same as that of a 50 keV X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is the energy equivalent of electron mass is 0.5 MeV)

The de-Broglie wavelength of an electron is the same as that of a 50 keV X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is ( the energy equivalent of electron mass of 0.5 MeV)

The de-Broglie wavelength of an electron is same as the wavelength of a photon. The energy of a photon is ‘x’ times the K.E. of the electron, then ‘x’ is: (m-mass of electron, h-Planck’s constant, c - velocity of light)

The de-Broglie wavelength of an electron is same as the wavelength of a photon. The energy of a photon is ‘x’ times the K.E. of the electron, then ‘x’ is: (m-mass of electron, h-Planck’s constant, c - velocity of light)

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

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.

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.

The de-Broglie wavelength of an electron is same as the wavelength of an photon.The K.E.of photon is 'x' times the K.E.of the electron then 'x' is (M- mass of electron h - Planck's constant C - Velocity of light)