Question

Copper(atomic mass $=63.5$) crystallizes in a FCC lattice and has density $8.93$g.$c{m}^{-3}$. The radius of copper atom is closest to:

• A. $361.6$ pm
• B. $511.4$ pm
• C. $127.8$ pm
• D. $102.8$ pm

Answer 1

Answer: (C)

$127.8$ pm

Theoretical density of a crystal is given by $\rho =\frac{nM}{N_A{a}^3}g/c{m}^3$

Given, $M=63.5g;\rho =8.93g/c{m}^3$

For FCC, $n=4$

$\Rightarrow a^3=\frac{4\times 63.5}{6.022\times {10}^{23}\times 8.93}$

$\Rightarrow a={\left(\frac{4\times 63.5}{6.022\times {10}^{23}\times 8.93}\right)}^{1/3}=3.6\times {10}^{-8}cm$

Now, if $r$ is radius of copper atom, then in FCC,

$r=\frac{a}{2\sqrt{2}}$

$r=\frac{3.6}{2\sqrt{2}}\times {10}^{-8}cm=127.8pm$