Question

Problem 4: X-Ray diffraction is a technique to determine the crystallographic structure of single crystal materials. The figure below shows the diffraction pattern of a aluminum. The first eight planes that x-ray diffraction peaks for aluminum are indicated in terms of their Miller indices. Sketch these planes and their intercepts relative to the cubic unit (draw the unit cell as it is shown below). (111) 100 ? = 0.1542 nm (CuK$_\alpha$-radiatio Intensity (arbitrary units) 80 60 (200) (220) (311) 40 20 (222) (400) (331)(420) 0 20 30 40 50 60 70 80 90 100 110 120 2$\theta$ (degrees) a y a Include the intercepts and Miller indices in each sketch

          Problem 4: X-Ray diffraction is a technique to determine the crystallographic structure of single
crystal materials. The figure below shows the diffraction pattern of a aluminum. The first eight
planes that x-ray diffraction peaks for aluminum are indicated in terms of their Miller indices.
Sketch these planes and their intercepts relative to the cubic unit (draw the unit cell as it is shown
below).
(111)
100
? = 0.1542 nm (CuK$_\alpha$-radiatio
Intensity (arbitrary units)
80
60
(200)
(220)
(311)
40
20
(222)
(400) (331)(420)
0
20 30 40 50 60 70 80 90 100 110 120
2$\theta$ (degrees)
a
y
a
Include the intercepts and Miller
indices in each sketch

$Problem 4: X-Ray diffraction is a technique to determine the crystallographic structure of single crystal materials. The figure below shows the diffraction pattern of a aluminum. The first eight planes that x-ray diffraction peaks for aluminum are indicated in terms of their Miller indices. Sketch these planes and their intercepts relative to the cubic unit (draw the unit cell as it is shown below). (111) 100 ? = 0.1542 nm (CuKα-radiatio Intensity (arbitrary units) 80 60 (200) (220) (311) 40 20 (222) (400) (331)(420) 0 20 30 40 50 60 70 80 90 100 110 120 2θ (degrees) a y a Include the intercepts and Miller indices in each sketch$

Added by Jill A.

Question

Please give Ace some feedback