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 of a photon of radiated light:

,

where and are the kinetic energies of the electron before and after radiation, respectively.
This kind of radiation is called bremsstrahlung radiation, 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 7.00 , what is the minimum wavelength of light radiated by such beam directed head-on into a lead wall?
Express your answer numerically in angstroms.

Explanation / Answer

Ke = 1/2 * m * v^2 = 1.6 * 10E-16 (1000 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 * 10E-16 lambda = 1.24 * 10E-9 m = 1.24 nm

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