If wavelength `(lamda)` of a photon and de Broglie wavelength of an electron are same, then show that the energy of the photon is `(2 lamda mc)/(h)` times the kinetic energy of the electron. Here, m, c and h have their usual meanings

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 of photon of lambda is equal to the de-Broglie wave length of electron. Show that energy of photon =2lambdamc/hxx kinetic energy of electron .

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 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 wavelength of an electromagnetic radiation is lamda . What is the energy of a photon of this radiation?

The de Broglie wavelength of a proton and an electron are same, then what is the relation between their kinetic energies?

Calculate the de Broglie wavelength of an electron of kinetic energy 500 eV.

The de Broglie wavelength of an electron and the wavelength of a photon are equal and its value is lamda= 10^(-10)m . Which one has a higher kinetic energy?

The de Broglie wave length of an electron of velocity 1.5×10^8 m/s is equal to that of photon. The ratio of the kinetic energy of electron and photon will be—