Question

Gold (atomic radius= 0.144 nm) crystallizes in fcc. The length of a side of the unit cell will be

Answer 1

Step 1: Finding the relation

For a face-centred unit cell (fcc),

$\mathrm{r}=\mathrm{a}÷2\surd 2$

$\mathrm{r}=$the radius of the unit cell

$\mathrm{a}=$length of the side of the unit cell

$\surd 2=1.414$

Step 2: Calculating the length of the side

Given the atomic radius of Gold is $0.144\mathrm{nm}$

The length of the side of a unit cell will be.

$\mathrm{r}=\mathrm{a}÷2\surd 2$

$\mathrm{a}=\mathrm{r}\times 2\surd 2$

$\mathrm{a}=0.144\times 2\times 1.414$

$\mathrm{a}=0.407\mathrm{nm}$

Therefore the length of the side of the unit cell is $0.407\mathrm{nm}$.