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.

Part A
Given an electron beam whose electrons have kinetic energy of 9.00 keV, what is the
minimum wavelength Lambdamin of light radiated by such beam directed head-on into a lead
wall?
Express your answer numerically in nanometers.

See Answers (1)

Suggested Answer

The correct answer is 1.3777 x 10 ^-2 m.What is bremsstrahlung radiation? Bremsstrahlung, also known as "braking radiation" or "deceleration radiation," is electromagnetic radiation that is produced when a charged particle decelerates when it is deflected by another charged particle. This is commonly an electron by an atomic nucleus.The energy of the photon is given by:E=hν=hcλE=hν=hcλHere• h=6.63⋅10^−34 J⋅s=6.63⋅10^−34 J⋅s is the Plank's constant and c is the speed of light and λ is the wavelength λ= hc/E = 4.136×10−15 eV·s (2.998x10^8 m/s)/10^4 eV = 4.136×10^−15 eV·s (2.998x10^8 m/s)/9^4 eV = 1.3777 x 10 ^ (8-15-4) = 1.3777 x 10 ^-11 m = 1.3777 x 10 ^-2 mTo learn more about bremsstrahlung radiation from the given linkhttps://brainly.com/question/25746629#SPJ1