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 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 4.00 keV, what is the minimum wavelength Amin of light radiated by such beam direct head-on into a lead wall? Express your answer numerically in nanometers. ► View Available Hint(s) 195] ΑΣΦ Amin 0.206 Submit Previous Answers ? nm
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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:
Eγ=K−K′��=�−�′,
where K� and K′�′ 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.
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