Question

Atomic radius of silver is 144.5 pm. The unit cell of silver is a face centred cube. Calculate the density of silver.

• A. 10.50 g/cm³
• B. 16.50 g/cm³
• C. 12.30 g/cm³
• D. 15.50 g/cm³

Answer 1

Answer: (A)

10.50 g/cm³

Answer 2

For (fcc) unit cell, atoms touch each other along the face diagonal.

Hence, Atomic radius (R) $= \frac{a\sqrt{2}}{4}$

$a = \frac{4R}{\sqrt{2}} = \frac{4}{\sqrt{2}} \times 144.5pm = 408.70pm = 408.70 \times 10^{- 10}cm$

Density (4) $= \frac{ZM}{VN_{0}},$ $V = a^{3}$

D = $\frac{ZM}{a^{3}N_{0}}$; where $Z$ for (fcc) unit cell = 4 , Avagadro’s number $(N_{0}) = 6.023 \times 10^{23}$, Volume of cube ($V$)$= (408.70 \times 10^{- 10})^{3}cm^{3}$ and M (Mol. wt.) of silver = 108,

D $= \frac{4 \times 108}{(408.70 \times 10^{- 10})^{3} \times 6.023 \times 10^{23}}$ $= 10.50g/cm^{3}$