With the help of a labeled diagram show that there are four octahedral voids per unit cell in a cubic close-packed structure.
With the help of a labeled diagram show that there are four octahedral voids per unit cell in a cubic close-packed structure.
A cubic close-packed structure contains one atom at each of eight corners of a unit cell and one atom at each of six faces which can be represented below
As we know any atom surrounded by six atoms (hard sphere) creates an octahedral void. in case of fcc body centre is surrounded by six identical atoms present at face centre hence, there is an octahedral void a body centre of each unit cell.
Beside the body centre, there is one octahedral void a centre of each of 12 edges as shown below
Since each void is shared by 4 unit cell. Therefore, the contribution of the octahedral void to each edge of a unit cell is 14
Number of octahedral void a centre of 12 edge =14×12=3
Number of the octahedral void at body centre =1
Therefore, total number of octahedral void at each ccp lattice = 3 + 1 = 4
A cubic close-packed structure contains one atom at each of eight corners of a unit cell and one atom at each of six faces which can be represented below
As we know any atom surrounded by six atoms (hard sphere) creates an octahedral void. in case of fcc body centre is surrounded by six identical atoms present at face centre hence, there is an octahedral void a body centre of each unit cell.
Beside the body centre, there is one octahedral void a centre of each of 12 edges as shown below
Since each void is shared by 4 unit cell. Therefore, the contribution of the octahedral void to each edge of a unit cell is 14
Number of octahedral void a centre of 12 edge =14×12=3
Number of the octahedral void at body centre =1
Therefore, total number of octahedral void at each ccp lattice = 3 + 1 = 4
