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 volume-fraction 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 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%
To answer these questions, we need to understand the concept of a simple cubic unit cell and how atoms are arranged within it.
(i) A simple cubic unit cell is the most basic type of unit cell in a crystal lattice structure. It consists of atoms arranged at the corners of a cube, with one atom per corner. To sketch one face of a simple cubic unit cell, draw a square with each corner representing an atom.
(ii) In the case where the maximum fraction of the lattice in volume is filled with atoms, each atom is approximated as a hard sphere. The radius of each atom can be determined by considering the length of the side of the unit cell (a). Since each atom is at a corner, the distance between the centre of the unit cell and the corner atom is equal to half the length of the diagonal of the face of the cube. Using Pythagoras' theorem, we can find this distance, which corresponds to the radius of the atom.
The diagonal length of the face of the cube can be calculated as: d = √(a^2 + a^2) = √2a.
Therefore, the radius of each atom is r = 0.5 * √2a.
(iii) The volume of each atom can be calculated using the formula for the volume of a sphere. The volume of a sphere is given by: V = (4/3) * π * r^3. Substitute the value of r from the previous step to obtain the volume of each atom in terms of a.
(iv) In a simple cubic unit cell, there are 8 corners, and each corner is occupied by one atom. Therefore, 8 atoms contribute volume to the unit cell.
(v) Each corner atom is shared by 8 unit cells. Only 1/8th of each corner atom is inside the unit cell.
(vi) The volume filled by all the corner atoms can be calculated by multiplying the volume of each individual atom by the number of atoms contributing to the unit cell. This would be:
Volume filled by corner atoms = 8 * (1/8) * (volume of each atom)
(vii) The total volume of the unit cell can be calculated by multiplying the length of each side, a, three times (since it is a cube). Therefore, the volume of the unit cell is given by: V = a * a * a = a^3.
(viii) The maximum fraction of the lattice volume filled with atoms can be calculated by dividing the volume filled by all the corner atoms (from step vi) by the volume of the unit cell (from step vii).