Question

The edge length of unit cell of a metal having molecular weight $75$ g/mol is $5\mathring{A}$
which crystallizes in cubic lattice. If the density is $2g/cc$, then find the radius (in pm) of metal atom $(N_A=6\times {10}^{23})$. Give answer in $pm$.

Answer 1

$d=\frac{Z\times M}{N_A\times (a\times {10}^{-8}{)}^3}$

$Z=\frac{2\times 6\times {10}^{23}\times (5\times {10}^{-8}{)}^3}{75}$

$Z=2$

$r=\frac{\sqrt{3}\times a}{4}$

$r=216.5pm$

Both 216 and 217 are correct answers.