0.259 52
. The fraction of empty space in a ccp cell is 0.259 52
Cubic closest packing is another name for <em>face centred cubic</em>.
Refer to the attached diagram of an fcc unit cell from <em>hemantmore</em>.
<h2>a) <em>
Volume of atoms in unit cell
</em></h2>
a. <em>Number of atoms in unit cell
</em>
In an fcc lattice, there are eight atoms at the eight corners of the cube and one at the centre of each face
∴ No. of atoms
= 8 corners x (1/8 atom)/(1 corner) + 6 faces x (½ atom/1 face)
= 1 atom + 3 atoms = 4 atoms
b. <em>Volume of atoms </em>
Let <em>r</em> be the radius of an atom. Then
Volume of 1 atom = (4/3)π<em>r</em>^3
Total volume of atoms = 4 atoms × [(4/3)π<em>r</em>^3]/(1 atom) = (16/3)π<em>r</em>^3
<h2>b) <em>
Volume of unit cell
</em></h2>
The length of the diagonal AB is
AB = 4<em>r</em>
Let <em>a</em> be the edge length of the cube.
AB^2 = AC^^2 + BC^2
(4<em>r</em>)^2 = <em>a</em>^2 + <em>a</em>^2
16<em>r</em>^2= 2<em>a</em>^2
<em>a</em>^2 = 8<em>r</em>^2
<em>a</em> = <em>r</em>x8^½
<em>V</em> = <em>a</em>^3 = (<em>r</em>x8^½)^3 = 8<em>r</em>^3x8^½
<h2>c) <em>
Fraction of occupied space</em> (packing efficiency, PE)
</h2>
PE = Volume of atoms/Volume of cell = [(16/3)π<em>r</em>^3]/[ 8<em>r</em>^3x8^½]
= 2π/(3x8^½)
<h2>d) <em>
Fraction of empty space
</em></h2>
Fraction of empty space = 1 - PE = 1 - 2π/(3x8^½) = 1 – 0.740 480
= 0.259 52