The unit cell of metallic gold is face-centred cubic.
(a) How many atoms occupy the gold unit cell?
(b) What is the mass of a gold unit cell?

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Solution

(a) gold crystallises in FCC lattice
$Z_{e f f e c t i v e}$ of gold= $8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$ atoms
(b) One unit cell of gold contains $4$ gold atoms
So, mass of gold unit cell= $4 \times M_{g o l d}$
= $4 \times 197 = 788$ $a m u$

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The unit cell of metallic gold is face-centred cubic.(a) How many atoms occupy the gold unit cell?(b) What is the mass of a gold unit cell?

(a) gold crystallises in FCC latticeZeffective of gold= 8×18+6×12=4 atoms(b) One unit cell of gold contains 4 gold atomsSo, mass of gold unit cell= 4×Mgold=4×197=788 amu

The unit cell of metallic gold is face-centred cubic.
(a) How many atoms occupy the gold unit cell?
(b) What is the mass of a gold unit cell?

(a) gold crystallises in FCC lattice
$Z_{e f f e c t i v e}$ of gold= $8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$ atoms
(b) One unit cell of gold contains $4$ gold atoms
So, mass of gold unit cell= $4 \times M_{g o l d}$
= $4 \times 197 = 788$ $a m u$