A relativistic electron has a de Broglie wavelength of 2.45 pm (1 pm = 10-¹2 m). Determine its velocity, expressed as a fraction of the speed of light c.

(a) 620 nanometers (nm)? (b) What are the energy and momentum of a photon of red light of wavelength What is the wavelength (in nm) of photons of energy 2.40 eV? 1.

Louis de Broglie proposed that a particle of mass m and moving at the speedv will have a wavelength related to its momentum. Air molecules of mass 30 amu travel at speeds which you can take to be 500 ms−1. Calculate their de Broglie wavelength.

In quantum interpretation of the electromagnetic waves in vacuum the photon has the energy E = hw/2n and the momentump= hk/2n. So the ratio E of the energy to the momentum is = c, the speed of light in k vacuum. Similar relation can be obtained in classical electrodynamics as follows. Consider the time-averaged energy density of the electromagnetic field ɛ = B² /2µ0 + €0 E² /2 for a plane wave propagating in vacuum along the z-direction. Calculate the time-averaged energy flux (Poynting flux) for such a wave S = E × H and confirm that its ratio to the time-averaged energy density ratio is equal to the speed of light in vacuum, S/e = c.