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

If the de broglie wavelength of an electron is 1Å, then the velocity of the electron will be

The de-broglie wavelength of a photon is twice the de-broglie wavelength of an electron. The speed of the electron is v_(e)=c/100 . Then

De-broglie's wavelength of an electron gas at 300K is x A^0 then x 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)

A photon sensitive metallic surface emits electrons when X-rays of wavelength lambda fal on it. The de-Broglie wavelength of the emitted electrons is (Neglect the work function of the surface, m is mass of the electron. H=Planck's constant. C= velocity of light)

The wavelenths gamma of a photon and the de-Broglie wavelength of an electron have the same value.Find the ratio of energy of photon to the kinetic energy of electron in terms of mass m, speed of light and planck constatn

The ratio of the wavelengths of a photon and that of an electron of same energy E will be [m is mass of electron ]