Question

Lithium has a $bcc$ structure. Its density is $530{\text{kg m}}^{-3}$ and its atomic mass is $6.94{\text{g mol}}^{-1}$. Calculate the edge length of a unit cell of lithium metal. $(N_A=6.02\times {10}^{23}{\text{mol}}^{-1})$

• A. $527\text{pm}$
• B. $264\text{pm}$
• C. $154\text{pm}$
• D. $352\text{pm}$

Answer 1

The correct option is D $352\text{pm}$

For bcc
$Z=2,\rho =530kg{m}^{-3}$, $\text{Mass of}Li=6.94gmo{l}^{-1},N_A=6.02\times {10}^{23}mo{l}^{-1}$
$\rho =530kg{m}^{-3}=\frac{530\times 1000g}{1\times (100{)}^{3}c{m}^3}=0.53gc{m}^{-3}$
$\rho =\frac{Z\times At.mass}{N_A\times a^3}$
$a^3=\frac{Z\times At.mass}{N_A\times \rho }=\frac{2\times 6.94}{6.02\times {10}^{23}\times 0.53}$
$=43.5\times {10}^{-24}c{m}^3$
$\Rightarrow a=352\times {10}^{-10}cm=352\text{pm}$