Unit – VIII

Radius Ratio of Interstitial Voids

  • The vacant space left in the closest pack arrangement of constituent particles is called interstitial void or interstitial site.

Tetrahedral Voids:

Locating Tetrahedral Voids:

  • Let us consider a unit cell of ccp or fcc lattice [Fig. (a)]. The unit cell is divided into eight small cubes. Each small cube has atoms at alternate corners [Fig. (b)]. In all, each small cube has 4 atoms. When joined to each other, they make a regular tetrahedron.
  • Thus, there is one tetrahedral void in each small cube and eight tetrahedral voids in total. Each of the eight small cubes have one void in one unit cell of ccp structure.
  • We know that ccp structure has 4 atoms per unit cell. Thus, the number of tetrahedral voids is twice the number of atoms.

Radius Ratio of Tetrahedral Void:

A tetrahedral site in a cube having tetrahedral void of radius r is as shown at the centre of the cube. Let R be the radius of the constituent particle of the unit cell. Let a be each side of the cube.

Consider face diagonal AB (Right angled triangle ABC). We have,



Along face diagonal AB the two sphere touch each other.

Consider body diagonal AD (Right angled triangle ABD). We have,

Along body diagonal AD the two sphere at the diagonals touch the tetrahedral void sphere..



Dividing equation (2) by (1)

Thus radius ratio for tetrahedral void is 0.225

Octahedral Voids:

Locating Octahedral Voids:

  • Let us consider a unit cell of ccp or fcc lattice [Fig. (a)]. The body centre of the cube, C is not occupied but it is surrounded by six atoms on face centres. If these face centres are joined, an octahedron is generated. Thus, this unit cell has one octahedral void at the body centre of the cube.
  • Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. [Fig. (b)]. It is surrounded by six atoms, four belonging to the same unit cell (2 on the corners and 2 on face centre) and two belonging to two adjacent unit cells.
  • Since each edge of the cube is shared between four adjacent unit cells, so is the octahedral void located on it. Only 1/4 th of each void belongs to a particular unit cell.
  • Thus in cubic close packed structure:
    Octahedral void at the body-centre of the cube = 1
    12 octahedral voids located at each edge and shared between four unit cells =12 x 1/4 = 3
    Thus, total number of octahedral voids 1 + 3 = 4
  • We know that in ccp structure, each unit cell has 4 atoms. Thus, the number of octahedral voids is equal to this number of atoms.

Radius Ratio of Octahedral Void:

A octahedral site in a cube having octahedral void of radius r is as shown at the centre of the cube. Let R be the radius of the constituent particle of the unit cell. Let a be each side of the cube. In this case we can see that a = 2R

Consider Right angled triangle ABC. We have,

Thus radius ratio for 0ctahedral void is 0.414



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3 Comments

  1. In radius ratio of tetrahedral voide why we have taken in eq.2 the radius of tetrahedral void 2r why not r ?

  2. In your radius ratio for tetrahedral void you state that the body diagonal AD includes the radius of two spheres. This is incorrect however as there is only one sphere on the body diagonal illustrated. This is further expressed by the fact that you yourself state when you say that each of the eight small cubes that contain this void only have atoms at alternating corners. By this logic it is impossible for a body diagonal and a face diagonal to start at the same corner AND both be intersecting an atom at each corner of the cube.. In fact, the corner you have labeled as D on your diagram is not the corner that the body diagonal goes to from A. The way you have labeled it, AD is not a body diagonal but a face diagonal. Because of this fundamental error in your set up, this answer seems incorrect to me. Yet, I do not see how to solve it without this faulty assumption so I will use this for my homework. Let me know if I am incorrect in my assessment.

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