Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

A deutron is accelerated through 100 V, its de Broglie wavelength is

The de Broglie wavelength of electron in second Bohr orbit is exactly equal to

The kinetic energy of an electron having de Broglie wavelength 2overset@A is

If the total energy of anelectron in a hydrogen like atom in excited state is -3.4eV ,then the de Broglie wavelength of the electron is

Write the expression for de-Broglie wavelength of a particle.