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 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.

Explanation / Answer

To find the minimum wavelength (or maximum energy) the electron must give up ALL of its energy. Therefore,

E (gamma) = KE ( electron ) = 6.00 keV.

The energy of a photon is equal to hc / ? where ? is the wavelength, c the speed of light, and h is Planck's constant. Plug in the energy (either convert to Joules or use Planck's Constant with units of eV-s) and solve for the wavelength.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.