Question

If the lattice parameter of Si is 5.43 ${\rm A}^\circ $ and the mass of Si atom is 28.08×1.66 ×${10^{ - 27}}$Kg, the density of silicon in kg \[{m^{ - 3}}\]is :
[Given: Silicon has a diamond cubic structure]
a. 2330
b. 1115
c. 3445
d. 1673

Answer 1

Hint: Density for a unit cell = $\dfrac{{Mass{\text{ }}of{\text{ }}one{\text{ }}unit{\text{ }}cell}}{{Volume{\text{ }}of{\text{ }}one{\text{ }}unit{\text{ }}cell}}$. As Si has diamond cubic structure, then its unit cell contains 8 atoms.

Complete answer:
First of all see what the values are given in the question.
Lattice parameter of Si= 5.43 ${\rm A}^\circ $
Mass of Si atom = 28.08×1.66 ×${10^{ - 27}}$Kg
We have to find out = density of silicon in Kg \[{m^{ - 3}}\].
To calculate the density of a unit cell we first have to know the mass of one unit cell by mass of one atom then the volume of the cubic cell of lattice.
Mass of one unit cell = no. of atom × mass of each atom
$= 8 × 28.08 \times 1.66 \times {10^{ - 27}}kg$
$= 3.73 \times 10^{ - 25}$ kg
Volume of one unit cell = (length of one side of unit cell) $^3$
$= (5.43×10^{ - 10}$)$^3$
$= 1.6×10^{ - 28}$ m$^3$
Now calculate the density of silicon by putting the values of mass of one unit cell and volume of one unit cell in the formula of density.
Density of silicon in kg m$^{ - 3}$ = $\dfrac{{Mass{\text{ }}of{\text{ }}one{\text{ }}unit{\text{ }}cell}}{{Volume{\text{ }}of{\text{ }}one{\text{ }}unit{\text{ }}cell}}$
=$\dfrac{{1.6 \times {{10}^{ - 28}}\;}}{{3.73 \times {{10}^{ - 25}}}}$ ​
=2330 kg m$^{ - 3}$

Hence, the correct option is (a) i.e. 2330 kg m$^{ - 3}$.

Note:
Unit conversion is given in the question otherwise if not given then Change the mass of atom given in a.m.u into kg by multiplying 1.66× ${10^{ - 27}}kg$to it for density into kg m$^{ - 3}$. Don’t get confused as the structure is given diamond cubic, this only provides the number of atoms in lattice.