Ques.The density of copper metal is 8.95 g cm³. 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 mol¹and N_A= 6.02 × 10²³mol¹)

can you please explain how it is fcc?d=M.z a³.N_AOn substituting we get,8.95=z.63.54 6.022X10²³X(1.27X10-8)³z=0.17.please help me out! where did i go wrong?

The density of metal is given as
$d = \frac{z\times M}{a^3\times N_A}$

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 ​$\times$ 10²³

For simple cubic,
$$\begin{gathered} d = \frac{1\times 63.54}{(2\times 127.8\times {10}^{-10}{)}^3 \times 6.02 \times {10}^{23}} \\ d = 6.31 g.c{m}^{-3} \end{gathered}$$

For BCC,
$$\begin{gathered} d = \frac{2 \times 63.54}{({\displaystyle \frac{4}{\sqrt{3}}}\times 127.8\times {10}^{-10}{)}^3 \times 6.02\times {10}^{23}} \\ d = 8.2 g.c{m}^{-3} \end{gathered}$$

For FCC,
$$\begin{gathered} d = \frac{4 \times 63.54}{({\displaystyle \frac{4}{\sqrt{2}}}\times 127.8\times {10}^{-10}{)}^3 \times 6.02\times {10}^{23}} \\ d = 8.92 g.c{m}^{-3} \end{gathered}$$

Given the density = 8.95 g.cm^-3 , which is nearest for FCC. Hence copper unit cell is face centered cubic.