Question

Copper (atomic mass $=63.5$) crystallises in a fee lattice and has density $8.93gc{m}^3.$ 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)

Given, density of copper lattice $=8.93gc{m}^{-}3$
Number of atoms in fee lattice, $Z=4$
As we know, $\rho =\frac{M\times N}{N_A\times {\alpha }^3}$
${\alpha }^3=\frac{\rho \times N_A}{M\times N}=\frac{8.93\times 6.023\times 12^{23}}{4\times 63.5}$
$\alpha ={\left(47.2\times 10^{-24}\right)}^{13}$
$\alpha =3.61\times 10^{-10}m$
$=361pm$
Also, in fee lattice
$\alpha =2\sqrt{2}R$
$r=\frac{\alpha }{2\sqrt{2}}=\frac{361}{2\sqrt{2}}=127.8pm$