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Define space lattice and basis. A metal crystallizes with a density of 2.7gm/cc and has a packing fraction of 0.74. Determine the mass of one atom if the nearest neighbour distance is 2.86Å.
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  • $Density\ =\ 2.7\ gm/cc, \ Packing\ fraction=0.74\ ,\ d=2.86 \ A^\circ \cdots (Given)$
  • $Convert\ 2.7\ gm\ /cc = 2700 \ kg/ m^3$
  • $As,\ Packing\ fraction=0.74\ (Hence,\ it\ is\ a\ FCC\ structure),$
  • $As,\ it\ is\ a\ FCC\ structure, \ Number\ of\ atoms\ per\ / unit\ cell\ = 4$
  • $As, d=2.86 A ^\circ for\ FCC,$

$d= \dfrac {\sqrt2 a} {4} = 2.86$

$a=2.86 \times\dfrac 4{\sqrt2 } = A ^\circ$

$Density ({\rho})= \dfrac {Mass} {Volume} = \dfrac{(Number\ of\ atoms / unit\ cell) \times (Volume\ of\ one\ atom)} {Volume\ of\ unit\ cell}\\$ ${\rho} = \dfrac{n (M / N) } {a ^3}\ $ $N\ is\ Avogadro's\ number=\ 6.023 \times 10^{23} (CGS) $ $M= \dfrac {a^3 \times N \times {\rho}} {n}$ $M= \dfrac {(3.302 \times 10 ^{-10})^3 \times (6.023 \times 10 ^{23}) \times {\ 2700}} {4}$ $M=\ 14.64 \times 10^{-3} \ kg = \ 14.64 \ g$ * $Mass\ of\ atom\ per\ unit\ cell\ = \dfrac {14.64 \ g} {4} = \ 3.66 \ gm$

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