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Metallic gold crystallizes in the face-centred cubic lattice. The edge length of the cubic unit cell, a = 4.070 A. Calculate the closest distance between gold atoms and the density of gold. Atomic mass of Au = 197 amu. |
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Answer» Solution :In a face-centred cubic CELL, radius `=(sqrt(2)a)/4`……………. (Eqn. 3) `therefore` the closed distance between two atoms = diameter `=2 xx (sqrt(2)a)/4 = a/sqrt(2)` `=(4.070)/sqrt(2) A = 2.878 A` Number of atoms in a face-centred unit cell = `8(1/8)+ 6(1/2)` Mass of 4 atoms PER unit cell `=4 xx 197` amu `=4 xx 197 xx (1.66 xx 10^(-24))g` `=1.308 xx 10^(-21) g` Volume of the unit cell `=a^(3)` `=(4.07 xx 10^(-8))^(3)` CC. `therefore` Density of gold `=(1.308 xx 10^(-21))/(4.07 xx 10^(-8))^(3) = 19.40 g//"cc"` |
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