Question

If the lattice parameter of $Si =5.43 \mathring{A}$ and the mass of $Si$ atom is $28.08 \times 1.66 \times 10^{-27} kg$, the density of silicon in $kg\, m ^{-3}$ is (Given: Silicon has diamond cubic structure)

• A. 2330
• B. 1115
• C. 3445
• D. 1673

Answer 1

Answer: (A)

$Si$ has fcc structure $\left(Z_{\text {eff }}=4\right.$ and $Si$ is also present in alternate TVs $(=4)$. So the total number of atom $=4+4=8$ /unit cell
Note: $Si$ has diamond cubic structure, refer section $1.17$
$\therefore \rho=\frac{Z_{ eff } \times A w}{N_{ A } \times a^3}\left[\text { Mass of $Si$ atom }=\frac{A w}{N_{ A }}\right] $
$=\frac{8 \times \text { Mass of Si atom }}{a^3} $
$=\frac{8 \times\left(28.08 \times 1.66 \times 10^{-27}\right) kg }{\left(5.43 \times 10^{-10}\right)^3 m ^3}$
${\left[1 \mathring{A}=10^{-10} m \right]}$
$ = 2330\,kg\,m^{-3}$