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k0ka [10]
3 years ago
5

Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the perc

entage of unit cell volume that is occupied in (a) a face- centered cubic lattice, (b) a body-centered cubic lattice, and (c) a diamond lattice.
Chemistry
1 answer:
tatiyna3 years ago
4 0

Answer:

  • The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05%
  • The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03%  
  • The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%

Explanation:

The percentage of unit cell volume = Volume of atoms/Volume of unit cell

Volume of sphere = \frac{4 }{3} \pi r^2

a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:

let the side of each cube = a

Volume of unit cell = Volume of cube = a³

Radius of atoms = \frac{a\sqrt{2} }{4}

Volume of each atom = \frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3 = \frac{\pi *a^3\sqrt{2}}{24}

Number of atoms/unit cell = 4

Total volume of the atoms = 4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6} = 0.7405

= 0.7405 X 100% = 74.05%

b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice

Radius of atoms = \frac{a\sqrt{3} }{4}

Volume of each atom =\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3 =\frac{\pi *a^3\sqrt{3}}{16}

Number of atoms/unit cell = 2

Total volume of the atoms = 2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8} = 0.6803

= 0.6803 X 100% = 68.03%

c) Percentage of unit cell volume occupied by atoms in a diamond lattice

Radius of atoms = \frac{a\sqrt{3} }{8}

Volume of each atom = \frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3 = \frac{\pi *a^3\sqrt{3}}{128}

Number of atoms/unit cell = 8

Total volume of the atoms = 8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16} = 0.3401

= 0.3401  X 100% = 34.01%

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