Question

The density of copper metal is $8.95$ gm $c{m}^{-3}$. If the radius of copper atom is $127.8$ pm, Will the copper unit cell be a simple cubic, a body-centred cubic or a face centred cubic? (At mass of Cu=$63.54$g $mo{l}^{-1}$ and N${}_A$ =$6.02\times {10}^{23}mo{l}^{-1}$)

Answer 1

Density of metal $\left(\rho \right)=\frac{Z\times M}{N_A\times a^3}$

For, simple cubic, $a=2r,Z=1$

$\therefore \rho =\frac{1\times 63.54}{(2\times 127.8\times {10}^{-10}{)}^3\times 6.02\times {10}^{23}}=6.31$ $g/c{m}^3$

For BCC, $Z=2,a=\frac{4}{\sqrt{3}}r$

$\therefore \rho =\frac{2\times 63.54}{{(\frac{4}{\sqrt{3}}\times 127.8\times {10}^{-10})}^3\times 6.02\times {10}^{23}}=8.2$ $g/c{m}^3$

For FCC, $Z=4,a=\frac{4}{\sqrt{2}}r$

$\therefore \rho =\frac{4\times 63.54}{{(\frac{4}{\sqrt{2}}\times 127.8\times {10}^{-10})}^3\times 6.02\times {10}^{23}}=8.92$ $g/c{m}^3$

The density of copper metal is $8.95$ $g/c{m}^3$ which is nearest to FCC. Therefore, the copper unit cell is face centre cubic.