Copper has a unit cell volume of (47.453 A^3) and a face centered unit cell. How would I go about finding the volume of atoms within the unit cell? Would you use Avogadro's # somehow?
I don't get the 47.453 A^3 for volume. That seems far too large (of course that depends upon what A stands for).
Whatever the volume, then
V1/3 = length of an edge of the unit cell = a. Then 4r = a(21/2).
When you have found r (radius of the atom), then volume of the atom = (4/3)*pi*r3.
4r = a(21/2) and
volume of the atom = (4/3)*pi*r3
I got it thank you!
The A^3 was suppose to stand for ångström
cubed...
what does the pi stand for?
To find the volume of atoms within the unit cell, you can use basic mathematical relationships and the concept of a face-centered unit cell.
In a face-centered unit cell, there are atoms at each of the eight corners and atoms at the center of each face. Since copper has a face-centered unit cell, it means that there are effectively four atoms located within each unit cell.
To calculate the volume of the atoms within the unit cell, you need to determine the volume occupied by each atom. This can be done by dividing the unit cell volume by the number of atoms. In this case, the unit cell volume is given as 47.453 ų and there are four atoms per unit cell (based on the face-centered structure).
So, the volume of each atom within the unit cell can be calculated as follows:
Volume of each atom = Unit cell volume / Number of atoms
= 47.453 ų / 4
The result will give you the volume occupied by each atom within the unit cell.
As for Avogadro's number (6.022 × 10^23 mol⁻¹), it is not directly used in this calculation since it represents the number of atoms or molecules in one mole of a substance. In this case, we are directly calculating the volume of atoms within a unit cell, so Avogadro's number is not necessary.
Keep in mind that this calculation assumes that each atom within the unit cell is of equal volume and the structure is ideal. In reality, the atomic size and shape might vary, and there could be interatomic spacing influences between neighboring unit cells.