Posted by ali on Sunday, February 19, 2012 at 9:26am.
. (i) Sketch one face of a simple cubic unit cell of side a for the case when the maximum fraction of the lattice in volume is filled with atoms, and each atom is approximated by a hard sphere. [5 marks]
(ii) What is the radius of each atom in terms of a? [4 marks]
a/2
(iii) What is the volume of each atom in terms of a? [4 marks]
(4/3)pi r3 = (4/3)pi (a/2)3
=(pi a3)/6
(iv) How many corner atoms contribute volume to the unit cell? [4 marks]
Corner atoms contribute volume to the unit cell is one from eight corners.
(v) What volume-fraction of each corner atom is inside the unit cell? [4 marks]
volume atom/volume unit cell
=[(pi a3)/6]/ a3
=pi/6
=0.5233 x 100% = 52.33%
(vi) In terms of a, how much of the volume of the unit cell is filled by all of the corner atoms? [5 marks]
Volume of the unit cell is filled by all of the corner atoms is due to its got 8 corners.
(vii) What is the total volume of the unit cell? [4 marks]
Total volume of the unit cell = a3
(viii) Calculate the maximum fraction of the lattice volume filled with atoms. [5 marks]
Maximum fraction of the lattice volume filled with atoms:
8 corners:
=[(pi a3)/6]/ 8a3
=pi/48
=0.0654 x 100% = 6.54%
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