In a class discussion, my teacher told me that 74.06% of the volume of each unit cell is occupied by atoms (in the cubic closest structure). Where does this percentage come from? How can I find it?
http://suraj.lums.edu.pk/~cmpe103w04/Assign1Solution.pdf
thank you!
The percentage you mentioned, 74.06%, represents the volume occupied by atoms in a cubic closest structure, also known as the face-centered cubic (FCC) crystal structure. This value is derived mathematically based on the arrangement of atoms in the unit cell.
To understand how this percentage is obtained, we need to consider the FCC crystal structure. In this structure, atoms are positioned at the corners and center of each face of the unit cell. This results in a total of four atoms per unit cell.
To find the volume occupied by atoms, we need to calculate the total volume of the unit cell and then determine the volume of the atoms within it. The formula for the volume of an FCC unit cell is given by:
Volume of unit cell = a³ * (4/3)
Where 'a' represents the length of each side of the unit cell.
Next, we calculate the volume of the atoms within the unit cell. Since there are four atoms per unit cell, we need to multiply the volume of a single atom by four. The volume of a single atom can be calculated using the formula:
Volume of atom = (4/3) * π * r³
Where 'r' represents the atomic radius.
Now, we can calculate the percentage of the unit cell volume occupied by the atoms. This is done by dividing the volume of the atoms by the volume of the unit cell, and then multiplying by 100:
Percentage = (Volume of atom / Volume of unit cell) * 100
By substituting the appropriate values for the atomic radius and the length of the unit cell into the formulas mentioned above, you can calculate the exact percentage.
It is important to note that the percentage may vary depending on the crystal structure and the arrangement of atoms within the unit cell. These calculations are specific to the FCC crystal structure.