what is the volume of the unit cell?
A metal crystallizes in a face-centered cubic lattice. The radius of the atom is 0.197 nm. The density of the element is 1.54 g/cm^3
And the question is?? And tell us what you don't understand about it.
The question is what is the volume of the unit cell?
I don't understand how to get the volume of it
The diagonal along the face of the crystal for a fcc is 4r. Since you know the radius you can calculate the length of the diagonal. Remember that the diagonal of a right triangle in which the sides are equal is a^2 + a^2 = (diagonal)^2 which allows you to calculate a, the edge of the cell. Then a^3 is the volume. Post your work if you get stuck.
To find the volume of the unit cell in a face-centered cubic (FCC) lattice, we'll need to know the side length of the unit cell.
In an FCC lattice, there are atoms at all eight corners of the cube and one atom at each face center, making a total of four atoms. The atom in the center of the unit cell is shared with eight neighboring unit cells.
To find the side length, we can start by determining the distance between two atoms along a face diagonal. In an FCC lattice, the face diagonal is equal to four times the radius of the atom (2√2r).
Given the radius of the atom is 0.197 nm, the face diagonal would be 4 x 0.197 nm = 0.788 nm.
Now, we can calculate the side length of the unit cell using the face diagonal. In an FCC lattice, the side length (a) can be found using the formula a = d/√2, where d is the face diagonal.
Plugging in the value of the face diagonal, we have a = 0.788 nm / √2.
Calculating this, we find that the side length of the unit cell (a) is approximately 0.556 nm.
To find the volume of the unit cell, we use the formula V = a^3, where a is the side length:
V = (0.556 nm)^3
Now, we need to convert nm to cm to match the density unit.
1 nm = 1 x 10^-7 cm
V = (0.556 x 10^-7 cm)^3
Evaluating this, we find that the volume of the unit cell is approximately 1.17 x 10^-22 cm^3.