Crystal_Lattice_Prob

# Crystal_Lattice_Prob - -23 cm 3 = 4.09 x 10-8 cm = l l = a...

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Chemistry 2B Workshop Learning Skills Center, UC Davis Jim Hollister Example of Crystal Lattice Problem Calculate the diameter of a silver, Ag, atom in pm. It crystallizes as a face-centered cubic (FCC) lattice and has a density of 10.5 g cm -1 at 300 K. 1. Calculate the volume of the unit cell. 4 atoms Ag 1 mol Ag 107.9g Ag 1 cm 3 = 6.83 x 10 -23 cm 3 (unit cell) 6.022 x 10 23 atoms Ag 1 mol Ag 10.5g Ag (unit cell) 2. Find the length, l , of the side of the unit cell (cube). (Some professors will label the side, a ) 3 6.83 x 10
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Unformatted text preview: -23 cm 3 = 4.09 x 10-8 cm = l ( l = a = side of cube) 3. Find the radius, r , using the appropriate equation relating the length of the cube to the radius of the atom. In this case, since silver is an FCC crystal, we use: l = 4 r so r = √ 2 l √ 2 4 r = √ 2 (4.09 x 10-8 cm) 1m 10 12 pm = 144 pm 4 100cm 1 m 2 r = diameter, d , so d = 2(144 pm) = 2.89 x 10 2 pm = 289 pm The temperature is not needed in the calculation; it is given to specify that the density is for that particular temperature....
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## This note was uploaded on 02/09/2010 for the course CHEM 35985 taught by Professor Osterloh during the Spring '10 term at UC Davis.

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