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 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 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 4 Å . What is its momentum ?

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)

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)

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)

The equivalent wavelength of a moving electron has the same value as that of a photon having an energy of 6xx10^(-17)J . Calculate the momentum of the electron.