Unformatted text preview: A monochromatic beam of X-rays with a wavelength of 0.1 nm is scattered by a gold metal foil. Scattered X-rays are observed at an angle of 60° from the direction of the incident beam. (a) What is the wavelength (in nm) of the scattered X-rays? (b) Find the shift in wavelength of the X-rays photons scattered at 60° from the direction of the incident beam. (c) Find the kinetic energy of the recoil electrons from the gold foil. (d) What is the maximum shift of the radiation scattered by the gold foil in this experiment? Solution:
Given: (a) Physics: λo = 0.1 nm and θ = 60° λ − λo = λc (1 − cos θ ) where λc =
h mc λo λ
e− θ λ = λo + λc (1 − cosθ ) = 0.1 nm + 2.43 (1 − cos 60°) × 10 −3 nm = 0.1012 nm
(b) (c) The shift is ∆λ = λ − λo = 0.0012 nm Based on the conservation of energy: K = E0 − E = hc λ0 − hc λ = 1240 eV .nm 1 1 − = 147.03 eV 0.1 nm 0.1012 nm cos θ = − 1 The maximum shift is for θ = 180° ∆λmax = λc (1 − cos 180°) = 2λc = 4.86 × 10 −3 nm ...
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This note was uploaded on 12/13/2010 for the course PHY 8 taught by Professor D during the Spring '10 term at MIT.
- Spring '10