Powder diffraction for a pure metal yields the following dhkl spacings values (nm): 0.2088, 0.1808, 0.1278,...

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Powder diffraction for a pure metal yields the following dhklspacings values (nm): 0.2088, 0.1808, 0.1278, 0.1090, 0.1044,0.09038, 0.08293, and 0.08083. A monochromatic x-radiation having awavelength of 0.1542 nm was used. Use this information to find thefollowing: a) Is the pure metal FCC or BCC? b) Determine thelattice parameter. (c) Determine the atomic radius and the metalusing the values listed in Table 3.1 (or from the list in the frontcover of Callister)

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3.6 Ratings (608 Votes)

d_hklÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â sin²Ï´=Î»²/4d_hkl²Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (sin²Ï´/0.04543)=h² + k² + l²Â Â Â Â Â Â Â Â Â Â

0.2088Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.1363Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 3

0.1808Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.1818Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 4

0.1278Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.3639Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 8

0.1090Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.5003Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 11

0.1044Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.5454Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 12

0.09038Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.7277Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 16

0.08293Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.8643Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 19

0.08083Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.9098Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 20

These correspond to hkl values of (111),(200),(220),(311) so on..

The reflection maxima thus observed from planes for which hkl are either all odd or all even. So the unit cell belongs to the FCC arrangement.

The larger Bragg angles can be measured more accurately than the smaller one, so more accurate value for sin²Ï´ is 0.9098.

For this the corresponding d_hkl value is 0.08083

So, d_hkl² = Î»²/4 sin²Ï´ = 4a²/h² + k² + l²

or, (0.08083² = 4a²/20

or, a = 0.1807 nm = 180.7 pm

c. Table is not given in the question.

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