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Maurinko [17]
3 years ago
11

Calculate the fraction of empty space in cubic closest packing to five significant figures.

Chemistry
1 answer:
nata0808 [166]3 years ago
4 0

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

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