Question

Copper crystallizes in face centred cubic lattice and has dencity of $8.930$g $c{m}^{-}3$ at $298$K. Calculate the radius of copper atom.(At mass of Cu=$63.55$u,N${}_A$=$6.02\times 1023mo{l}^{-}1)$.

Answer 1

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

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

$\therefore a^3=\frac{Z\times M}{N_A\times \rho }$

$\therefore a^3=\frac{4\times 63.55}{6.02\times {10}^{23}\times 8.930}$

$\therefore a^3=4.728\times {10}^{-23}$ $cm$

$\therefore (2\sqrt{2}r{)}^3=4.728\times {10}^{-23}$ $cm$

$\therefore 2\sqrt{2}r=3.616\times {10}^{-8}$ $cm$

$\therefore r=1.278\times {10}^{-8}$ $cm$

$\therefore r=1.278$ $\stackrel{\circ }{A}$

Therefore, the radius of copper atom is $1.278$ $\stackrel{\circ }{A}$