Asked by Ahmed
Calculate the radius of a palladium (Pd) atom, given that Pd has an FCC crystal structure, a density of 12.0 g/cm3 , and an atomic weight of 106.4 g/mol.
Answers
Answered by
DrBob222
The FCC means 4 atoms/unit cell.
106.4 g/mol x (1/6.022E23) = 1.77E-22 g
1.77E-22 x 4 atoms/unit cell = 7.07E-22 grams/unit cell
density of unit cell = 12.0 g/cc = mass/volume so
volume = 7.07 E-22 g/12.0 g/cc = 5.89E-23 cc. Since volume a^3 so a^1/3 so (5.89E-23)^1/3 = a = 3.92E-8 cm
r = 1.414*a/4 = ?
106.4 g/mol x (1/6.022E23) = 1.77E-22 g
1.77E-22 x 4 atoms/unit cell = 7.07E-22 grams/unit cell
density of unit cell = 12.0 g/cc = mass/volume so
volume = 7.07 E-22 g/12.0 g/cc = 5.89E-23 cc. Since volume a^3 so a^1/3 so (5.89E-23)^1/3 = a = 3.92E-8 cm
r = 1.414*a/4 = ?
Answered by
Benjamin
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Answered by
Benjamin
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