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$\mathrm{Al}$ (at. wt 27 ) crystallizes in the cubic system with a cell edge of $4.05 Ã…$. Its density is $2.7 \mathrm{~g}$ per $\mathrm{cm}^{3}$. Determine the unit cell type calculate the radius of the $\mathrm{Al}$ atom
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Verified Answer
The correct answer is:
$\mathrm{fcc}, 1.432 Ã$
$\rho=\frac{Z \times M}{N_{0} \times a^{3}}$,
$$
2.7=\frac{Z \times 27}{6.02 \times 10^{23} \times(4.05)^{3} \times 10^{-24}}
$$
$\therefore Z=4$
Hence it is face centred cubic unit lattice.
Again $4 r=a \sqrt{2}=5.727 Ã…$
$\therefore \mathrm{r}=1.432 Ã…$
$$
2.7=\frac{Z \times 27}{6.02 \times 10^{23} \times(4.05)^{3} \times 10^{-24}}
$$
$\therefore Z=4$
Hence it is face centred cubic unit lattice.
Again $4 r=a \sqrt{2}=5.727 Ã…$
$\therefore \mathrm{r}=1.432 Ã…$
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