Question

An element has a face centered cubic (fcc) structure with a cell edge of $200\text{pm}$. The atomic radius is:

• A. $\frac{\sqrt{3}}{2}\times 200\text{pm}$
• B. $\frac{1}{\sqrt{2}}\times 200\text{pm}$
• C. $\frac{1}{2\sqrt{2}}\times 200\text{pm}$
• D. $\frac{2}{\sqrt{3}}\times 200\text{pm}$

Answer 1

The correct option is C $\frac{1}{2\sqrt{2}}\times 200\text{pm}$

For FCC lattice:
$A{C}^2=A{B}^2+B{C}^2$
$(r+2r+r{)}^2=a^2+a^2$
$(4r{)}^2=2a^2$
$4r=\sqrt{2}a$
$\therefore r=\frac{\sqrt{2}a}{4}=\frac{\sqrt{2}\times 200}{4}$
$=\frac{1}{2\sqrt{2}}\times 200\text{pm}$