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Tutorial 1 Calculate the fraction of the volume of the unit cell actually occupied by atoms in the face centred cubic structure. For the fcc structure, the atoms touch along the face diagonal, so 4 r  2a  r  4  a  Va    3 2 2 a 2 2 then 3 There are 4 atoms per cubic unit cell So the fract...

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Tutorial 1 Calculate the fraction of the volume of the unit cell actually occupied by atoms in the face centred cubic structure. For the fcc structure, the atoms touch along the face diagonal, so 4 r  2a  r  4  a  Va    3 2 2

a 2 2

then

3

There are 4 atoms per cubic unit cell So the fraction of the total volume occupied by the atoms is 3

Va 4  a  1 16    4  0.740 3 Vc 3 2 2 a 48 2

 

Copper has a density of 8.885 x 103 kg/m3, a relative atomic mass of 63.57 and has a face centred cubic structure. Estimate: (a) How many moles of copper there are in 1 m3 (b) How many atoms are there in 1 m3 (c) The size of the unit cube (d) The atomic radius of copper in the solid state (e) The mass of the copper atom. mass of 63.57 gm = 0.06357 kg. Therefore 1m3 of copper contains 8.885 103 1.40 10 5 mole . 0.06357

(b) 1 mole = NA atoms, where NA = Avogadro’s number. So from part (a) we find that the number of atoms in 1 m3 is 1.40 10 5 mole N A 1.40 105 mole 6.022 1023 atoms.mole 1  8.42 1028 atoms

(c) Copper has an fcc structure with 4 atoms per unit cell density    Vc 

Mass of unit cell 4A Cu  Volume of Unit cell VC N A

4A Cu 4 63.57gm.mole  1   m 3 4.752 10 29 m 3 a 3 1 23 N  A 8.885gm 6.02210 mole

so the lattice constant is a 3 Vc  3 4.75210 29 m 3  0.362nm

(d) For an fcc crystal we have 4r  2a  r  so rCu

0.362nm   0.128nm 2 2

a 2 2

Determine the Miller indices for the planes shown in the following unit cells. z

1/3

z

1/2

y

x

x Looking at the first plane we find Intercepts Intercepts in terms of a b and c Reciprocals of Intercepts Reduction Thus a (230) plane

y

a/2 1/2 2 not necessary

b/3 1/3 3

Repeating this analysis for the second plane we find Intercepts a b Intercepts in terms of a b and c -1 -1 Reciprocals of Intercepts -1 -1 Reduction not necessary Thus a ( 1 1 1) plane

c  0

-c 1 1

Calculate the radius of a Palladium atom, given that Pd has an FCC crystal structure, a density of 12.00 x 103 kg m-3, and an atomic weight of 0.1064 kg/mole For an FCC system there are four atoms per unit cell. The following figure shows a cross section through one of the faces.

From this figure we see that a2 + a2 = (4R)2. Solving for the lattice constant a we have a  2R 2

the volume of the unit cell Vc, in terms of the atom radius is then



VC a 3  2R 2



3

3 16R 2

Then the density is Mass of unit cell 4A Pd 4A Pd density     Volume of Unit cell VC N A 16R 3 2 N A Solving for R from the above expression   4 APd R    16  2 N A 

13

   4atoms / cell 0.1064kg / mole   3 3 23 2 6.023 10 atoms / mole   16 1210 kg.m



 

 0.13810

9

13



m  0.138nm

Determine the indices for the directions shown in the following unit cell

(A) Looking at the coordinates of the start (x1, y1, z1) and end (x2, y2, z2) points of the vector A we have (x1, y1, z1) = (0, 0, 2/3) and (x2, y2, z2) = (1, 1, 1/3). The vector joining these two points is

(x 2  x 1) xˆ  ( y 2  y1 ) yˆ  ( z 2  z1 )ˆz 1  0xˆ  (1  0) yˆ  (1 / 3  2 / 3) ˆz xˆ  yˆ  1 / 3zˆ

Rationalizing to whole numbers gives ˆ ˆ  3y ˆ  1z 3x

so this is the [33 1 ] direction. Repeating this analysis for the other vectors we find B = [ 40 3 ] direction C = [ 3 61] direction and D = [1 1 1]...