Vanadium crystalizes in a body centered cubic lattice (the V atoms occupy only the lattice points) how many V atoms are present in a unit cell?
I tried to find a good site for you to read but failed. The statement the problem makes is confusing. While it may be true that the V atoms occupy only the lattice POINTS, it also is true that if it is body centered there must b an atom at the center. There are 8 corner V atoms shared with 8 neighbors; therefore, this counts as (1/8)*8 = 1 atom.
That + the atom in the center makes 2 V atoms in a unit cell for a bcc.
thank you!
To determine the number of Vanadium (V) atoms present in a body-centered cubic (BCC) lattice, we need to consider the arrangement of the atoms in a unit cell.
A BCC lattice consists of atoms located at each corner of the unit cell and one atom positioned at the center of the cell. Each corner atom is shared by eight adjacent cells, while the center atom belongs solely to the unit cell.
In a BCC structure, there are two atoms per unit cell: one atom at the center and one eighth of an atom at each corner. Since an eight of an atom is shared by eight cells, it contributes to a whole atom in one unit cell.
Therefore, the total number of Vanadium atoms in a unit cell of a BCC lattice is one whole atom from the center plus one whole atom from the corner, resulting in a sum of two Vanadium (V) atoms per unit cell for Vanadium crystalizing in a BCC lattice.