Given :-

1. Cu is FCC structure

2. interplanar distance (d) = 2.08 Aº

3. (hkl) = (111)

4. Atomic weight (A) = 63.54

    

To find :-

1. Density ($ { \rho}$)

2. Diameter

    

Formula:

$d =\dfrac {a}{\sqrt{h^2+k^2+l^2}}$

$2.08 = \dfrac{a}{ \sqrt{1^2+1^2+1^2}}$

$a = {2.08 \sqrt{3}}$ Aº

a = 3.603 Aº

As it is FCC structure (n = 4)

$a^3\rho = {n}$$({A\over N})$ $\rho = {n\over a^3}{( {A\over N})}$ $\rho = {4\over ( \sqrt {3}\times2.08\times10^{-10})^3}$${\times ({63.54\over6.023\times10^{26}})}$    $ = ({8.55\times10^{28} ) \times (1.05\times10^{-25}) }$ $Density(\rho)= {9.02 \times 10^3}$ For FCC  $r= {a \sqrt{2} \over 4}$ Diameter (2r) = ${a \sqrt{2} \over 2}$

Diameter = 2.54 Aº