Ques.The density of copper metal is 8.95 g cm^{3}. If the radius of copper atom is 127.8 pm, is the copper unit cell a simple cubic, a body-centred cubic or a face centred cubic structure? (Given: At. Mass of Cu = 63.54 g mol^{1}and N_{A}= 6.02 × 10^{23}mol^{1})

__M.z__a

^{3}.N

_{A}On substituting we get,8.95=

__z.63.54__6.022X10

^{23}X(1.27X10-8)

^{3}z=0.17.please help me out! where did i go wrong?

$d=\frac{z\times M}{{a}^{3}\times {N}_{A}}$

For Simple cubic,

z = 1 and a = 2r

For BCC,

z = 2 and $a=\frac{4}{\sqrt{3}}r$

For FCC,

z = 4 and $a=\frac{4}{\sqrt{2}}r$

Substituting values of z and a in density formula and calculating density;

Given that-

Mass, M = 63.54 g/mol

Radius, r = 127.8 $\times $ 10

^{-10}cm

Avagadro's number, N

_{A}= 6.02 $\times $ 10

^{23}

For simple cubic,

$d=\frac{1\times 63.54}{(2\times 127.8\times {10}^{-10}{)}^{3}\times 6.02\times {10}^{23}}\phantom{\rule{0ex}{0ex}}d=6.31g.c{m}^{-3}$

For BCC,

$d=\frac{2\times 63.54}{({\displaystyle \frac{4}{\sqrt{3}}}\times 127.8\times {10}^{-10}{)}^{3}\times 6.02\times {10}^{23}}\phantom{\rule{0ex}{0ex}}d=8.2g.c{m}^{-3}$

For FCC,

$d=\frac{4\times 63.54}{({\displaystyle \frac{4}{\sqrt{2}}}\times 127.8\times {10}^{-10}{)}^{3}\times 6.02\times {10}^{23}}\phantom{\rule{0ex}{0ex}}d=8.92g.c{m}^{-3}$

Given the density = 8.95 g.cm

^{-3}, which is nearest for FCC. Hence copper unit cell is face centered cubic.

**
**