When a fast electron (i.e., one moving at a relativistic speed) passes by a heav

Question

When a fast electron (i.e., one moving at a relativistic speed) passes by a heavy atom, it interacts with the atom's electric field. As a result, the electron's kinetic energy is reduced; the electron slows down. In the meantime, a photon of light is emitted. The kinetic energy lost by the electron equals the energy E? of a photon of radiated light:

E?=K?K?,

where K and K? are the kinetic energies of the electron before and after radiation, respectively.

This kind of radiation is called bremsstrahlungradiation, which in German means "braking radiation" or "deceleration radiation." The highest energy of a radiated photon corresponds to the moment when the electron is completely stopped

Given an electron beam whose electrons have kinetic energy of 6.00keV , what is the minimum wavelength ?min of light radiated by such beam directed head-on into a lead wall?

Explanation / Answer

Ke = 1/2 * m * v^2 = 1.6 * 10E-16 (6000 eV)
If all of the energy is converted into radiation then
E = h f = h c / lambda where f is the radiated frequency
lambda = h c / E = 6.62 * 10E-34 * 3 * 10E8 / 1.6 x 6000* 10E-16
lambda = 0.2067 * 10E-9 m = 0.2067 nm

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