Question
Question: If the lattice parameter of \( Si \) is \( 5.43A^\circ \) and the mass of atom \( Si \) is \( 28.08 ...
If the lattice parameter of Si is 5.43A∘ and the mass of atom Si is 28.08×1.66×10−27kg , the density of silicon in kgm−3 is:
[Given: Silicon has a diamond cubic structure.]
(a) 2330
(b) 1115
(c) 3445
(d) 1673
Solution
The silicon is a chemical compound having symbol Si and atomic number 14 . The silicon is a tetravalent, metalloid and semiconductor (Pure Silicon is needed for these purposes) in computers and microelectronics.
Density of the atom,
d=Vm
Where, d is the density, V is the volume and m is the mass of the compound.
Volume of the cube,
V=l3
Where, l is the side of the cube.
Complete step-by-step answer:
First of all, we are writing the data given in the question.
Given data:
Mass of atom =28.08×1.66×10−27
Lattice parameter of the atom =5.43A∘=5.43×10−10
Lattice parameter of the atom is the measure of the side of the cube.
Now, putting the value of side as the lattice parameter of the atom and finding the volume of the cube-
So, the volume of the atom is 1.6×10−28m3 .
Now, we have to find the mass of the unit cell.
As in the question given that the Silicon atom is having a cubic structure. The number of atoms in the unit cells in the cubic structure is eight.
m=8×massofoneatom=8×28.08×1.66×10−27kg=3.73×10−25kg
So, the calculated mass of the unit cell is 3.73×10−25kg .
Now, putting the value of mass and volume in the density formula as given in the question.
d=Vm=1.6×10−283.73×10−25=2330kgm−3
So, the density of the atom will be 2330kgm−3 .
Hence, the correct option is (a) 2330 .
Additional Information:
Density is mass per unit volume. It is an intensive property (The intensive property is the property of matter that does not change with the amount of matter).
Note:
The cubic crystal system is the system in which the shape of the unit cells are cubic in structure and the unit cells are the smallest repeating unit having the total symmetry of the crystal structure.