Given: Metallic radius of aluminium
=125 pm=125×10−12 m
Length of the side of the unit cell
We know, for cubic close-packed structure:
a=2√2 r ............ (1)
Where, ′r′ is the radius of sphere and ′a′ be the edge length of the cube.
Substituting the value of ′r′ in equation (1), we get
a=2×1.414×125=353.55 pm≈354 pm
Hence, length of the side of the unit cell is 354 pm.
Final answer: Length of the side of the unit cell is 354 pm.
Given: Metallic radius of aluminium
=125 pm=125×10−12 m
Number of unit cells
Volume of one unit cell =a3=(354 pm)3
=4.4×107 pm3
[1 pm=10−12 m=10−10 cm]
=4.4×107×10−30 cm3
=4.4×10−23 cm3
Number of unit cells =Total volume of unit cellvolume of one unit cell
Therefore, number of unit cells in 1.00 cm3
=14.4×10−23=2.27×1022
Hence, number of the unit cells in 1.00 cm3 of aluminium are 2.27×1022.
Final answer: Number of the unit cells in 1.00 cm3 of aluminum are 2.27×1022.