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 $(Z_{\text{eff }}=4$ 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 $1.17$
$\therefore \rho =\frac{Z_{eff}\times Aw}{N_A\times a^3}\left[\text{ Mass of Si atom }=\frac{Aw}{N_A}\right]$
$=\frac{8\times \text{ Mass of Si atom }}{a^3}$
$=\frac{8\times \left(28.08\times 1.66\times 10^{-27}\right)kg}{{\left(5.43\times 10^{-10}\right)}^3{m}^3}$
$\left[1\mathring{A}=10^{-10}m\right]$
$=2330kg{m}^{-3}$