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.
To understand and answer these questions, we will need to consider the concept of a simple cubic unit cell in a crystal lattice with atoms approximated as hard spheres. Let's go through each question step by step:
(ii) The radius of each atom in terms of 'a':
In a simple cubic unit cell, the atoms are located at the corners. Since each atom is approximated as a hard sphere, it will touch the neighboring atoms along the edge of the cube. In this case, the diameter of the atom will be equal to the edge length 'a' of the unit cell. So, the radius of each atom can be expressed as half the edge length, which is 'a/2'.
(iii) The volume of each atom in terms of 'a':
The volume of a sphere is calculated using the formula V = (4/3) * π * r³, where 'r' is the radius of the sphere. Substituting the radius from the previous point, we get V = (4/3) * π * (a/2)³ = (1/6) * π * a³. Therefore, the volume of each atom is (1/6) * π * a³.
(iv) Number of corner atoms contributing volume to the unit cell:
In a simple cubic unit cell, each corner of the cube is occupied by an atom. Therefore, there are eight corner atoms in total that contribute to the volume of the unit cell.
(v) Volume fraction of each corner atom inside the unit cell:
Since each corner atom is shared by eight unit cells, only 1/8th of the volume of each atom is inside the unit cell. This is because the atom is located at the corner and is shared equally by eight neighboring unit cells.
(vi) Volume of the unit cell filled by all corner atoms in terms of 'a':
As mentioned in the previous point, each corner atom contributes only 1/8th of its volume to the unit cell. Therefore, the total volume filled by all the corner atoms can be calculated as (1/8) * 8 * (1/6) * π * a³ = (1/6) * π * a³.
(vii) Total volume of the unit cell:
In a simple cubic unit cell, the unit cell is in the shape of a cube, and all the atoms are located at the corners. The volume of a cube is calculated as V = a³, where 'a' is the edge length of the cube. Therefore, the total volume of the unit cell is a³.
(viii) Maximum fraction of the lattice volume filled with atoms:
To calculate the maximum fraction of the lattice volume filled with atoms, we need to divide the total volume filled by all corner atoms (from point vi) by the total volume of the unit cell (from point vii). This can be represented as [(1/6) * π * a³] / a³ = (1/6) * π. Therefore, the maximum fraction of the lattice volume filled with atoms is (1/6) * π.
I hope this helps in understanding the concepts and calculations involved in a simple cubic unit cell with atoms approximated as hard spheres.