The body-centered cubic unit cell of a particular crystalline form of silver is 2.2887A on each side. Calculate the density of this form of silver?
To calculate the density of a crystal structure, we need to know the number of atoms in the unit cell and the atomic weight of the element.
In the case of a body-centered cubic (BCC) unit cell, there is one atom at each corner and one atom at the center of the cube. So, the number of atoms per unit cell is 2.
The atomic weight of silver (Ag) is 107.87 g/mol.
Now, we can calculate the volume of the unit cell. Since it is a cube, the volume can be found by raising the length of one side to the power of 3:
Volume (V) = (Side length)^3
V = (2.2887A)^3 = 11.634 A^3
Next, we need to convert the volume from cubic Angstroms to cubic centimeters, as the atomic weight is given in grams per mole and density is typically expressed in grams per cubic centimeter (g/cm^3):
1 A^3 = 1 × 10^-24 cm^3
So, V = 11.634 × 10^-24 cm^3
Now, we can calculate the density:
Density (ρ) = Mass (m) / Volume (V)
Since we know the number of atoms per unit cell (n = 2) and the atomic weight (AW = 107.87 g/mol), the mass of the unit cell can be calculated as:
Mass = n × AW / Avogadro's number
Avogadro's number (NA) = 6.022 × 10^23
Mass = 2 × (107.87 g/mol) / (6.022 × 10^23 atoms/mol)
Finally, we can calculate the density:
Density = Mass / Volume
Plug in the values:
Density = (2 × 107.87 g/mol) / (6.022 × 10^23 atoms/mol) / (11.634 × 10^-24 cm^3)
Calculating this expression gives us the density of the crystalline form of silver.