CCP or FCC has two lattice point corner as well as face centred:

Suppose
‘r’ be the radius of sphere and ‘a’ be the edge length of the cube As there are
4 sphere in FCC unit cell

**(1) Relation between radius (r) and side (a)**

In FCC,
the corner spheres are in touch with the face centred sphere. Therefore, face
diagonal AD is equal to four times the radius of sphere AC = 4r

But from
the right angled triangle ABC:

**(2)**

**Effective no. of atoms per unit cell (Z):**

**(3) Volume of four spheres (atoms):**

**(4) Volume of unit cube:**

**(5) Packing efficiency (PE):**Percentage of space occupied by sphere

**(6) Percentage Voids:**100- PE= 26 %

**(7) Density of FCC(CCP):**

**(8) Coordination Numbers:**

**(9) LOCATION OF VOIDS: FCC/CCP UNIT CELL:**

**(**

**A) Tetrahedral voids**

**:**

**The FCC/CCP unit cell has**

**eight**tetrahedral voids per unit cell. Just below every corner of the unit cell, there is one

**tetrahedral void.**As there are eight corners, there are

**eight tetrahedral**voids.

**(**

**B) Octahedral voids**

**:**

**In an FCC/CCP unit cell, there are**

**four**octahedral voids. They are present at

**all the edge centres**and at the

**body centre.**The contribution of the edge centre is 1/4

Hence, total number of octahedral
voids:

**In CCP/FCC:**

Rank (Z) = 4,

Number of tetrahedral voids = 8 and

Number
of tetrahedral voids = 2 × Z

Number of tetrahedral voids in close
packing = 2 × eff. no. of spheres.

**Hence, there are two Tetrahedral Voids per sphere in closed packing arrangements.**

**In CCP/FCC:**

Z = 4

Number of octahedral voids = 4

Number of octahedral voids = Z

**There is exactly one OV per sphere in close packing.**

## No comments:

## Post a Comment