Solve - An old midterm question 3.67 - Determine the expected diffraction angle for the first-order reflection from the (310) set of planes for BCC chromium (Cr) when monochron latic radiation of wavelength 0.0711 nm is used. (Hint: Bragg's law)

To determine the expected diffraction angle for the first-order reflection from the (310) set of planes for BCC chromium (Cr) when monochromatic radiation of wavelength 0.0711 nm is used, we can use Bragg's law.<br /><br />Bragg's law states that the condition for constructive interference (diffraction) of X-rays by a crystal lattice is given by:<br /><br />nλ = 2d sin θ<br /><br />where n is the order of diffraction, λ is the wavelength of the incident radiation, d is the interplanar spacing of the crystal lattice, and θ is the diffraction angle.<br /><br />For the first-order reflection, n = 1. The interplanar spacing d for the (310) set of planes can be calculated using the lattice constants a of the BCC crystal structure. For BCC chromium, a = 0.287 nm. The interplanar spacing d for the (310) set of planes is given by:<br /><br />d = a / sqrt(310^2 + 310^2 + 310^2)<br /><br />Substituting the values, we get:<br /><br />d = 0.287 / sqrt(310^2 + 310^2 + 310^2) nm<br /><br />Now, we can use Bragg's law to calculate the diffraction angle θ:<br /><br />1λ = 2d sin θ<br /><br />Substituting the values, we get:<br /><br />1/0.0711 = 2(0.287 / sqrt(310^2 + 310^2 + 310^2)) sin θ<br /><br />Solving for θ, we get:<br /><br />θ = arcsin((2 * 0.287) / (sqrt(310^2 + 310^2 + 310^2) * 0.0711)) nm<br /><br />θ ≈ 28.8 degrees<br /><br />Therefore, the expected diffraction angle for the first-order reflection from the (310) set of planes for BCC chromium (Cr) when monochromatic radiation of wavelength 0.0711 nm is used is approximately 28.8 degrees.