In AB...AB...AB type arrangement if all the atoms present in body are removed,then packing fraction of the crystal is?
To determine the packing fraction of a crystal in an AB...AB...AB type arrangement, you need to consider the arrangement of atoms and the inherent formula unit of the crystal structure.
In this type of arrangement, each atom has two atoms of the other type as nearest neighbors. Let's consider the arrangement of atoms as follows:
A B A B A B ...
A B A B A B ...
Here, A and B represent different types of atoms. In this arrangement, each atom is surrounded by two atoms of the other type. Each atom is also in contact with six surrounding atoms, three of its own type and three of the other type.
To calculate the packing fraction, we need to determine the volume occupied by the atoms and the total volume of the unit cell.
The volume of one atom can be approximated as a sphere, and the volume of that sphere can be expressed as V_atom = (4/3)πr^3, where r is the atomic radius.
The total volume occupied by the atoms can be calculated by multiplying the volume of one atom by the number of atoms per unit cell. In this case, since there is one atom per unit cell for each type (A and B), the total volume occupied by the atoms will be 2∙V_atom.
The total volume of the unit cell can be calculated by using the edge length (a) of the unit cell. Assuming that the atoms are close-packed spheres, the total volume of the unit cell can be expressed as V_unitcell = a^3.
Finally, to calculate the packing fraction (η), divide the total volume occupied by the atoms by the total volume of the unit cell and multiply by 100:
η = (2∙V_atom / V_unitcell) ∙ 100
This formula will give you the packing fraction of the crystal in the AB...AB...AB type arrangement.
Keep in mind that this calculation assumes idealized close-packing, and in real crystals, there may be defects and other factors that can affect the packing fraction.