chemistry
posted by ali on .
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.
(ii) What is the radius of each atom in terms of a?
(iii) What is the volume of each atom in terms of a?
(iv) How many corner atoms contribute volume to the unit cell?
(v) What volumefraction of each corner atom is inside the unit cell?
(vi) In terms of a, how much of the volume of the unit cell is filled by all of the corner
atoms?
(vii) What is the total volume of the unit cell?
(viii) Calculate the maximum fraction of the lattice volume filled with atoms.

. (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 volumefraction 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%