PACKING FRACTIONS

Both of the above patterns of packing (i.e. hcp & ccp) though different in form are equally efficient. They occupy the maximum possible space which is about 74% of the available volume. Hence they are called closest packing.

In addition to the above two types of arrangements a third type of arrangement found in metals is body centred cubic (bcc) in which space occupied is about 68%.

CALCULATION OF THE SPACE OCCUPIED

In simple cubic unit cell       

Let ‘a’ be the edge length of the unit cell and r be the radius of sphere.

As sphere are touching each other

Therefore a = 2r

No. of spheres per unit cell = 1/8 × 8 = 1 

Volume of the sphere = 4/3 πr3 

 

1965_cubic unit cell.JPG
 

Volume of the cube = a3= (2r)3 = 8r3

∴ Fraction of the space occupied = 1/3πr3 / 8r3 = 0.524

∴ % occupied = 52.4 %

 

In face centred unit cell

Let ‘r’ be the radius of sphere and ‘a’ be the edge length of the cube

As there are 4 sphere in fcc unit cell

∴ Volume of four spheres = 4 (4/3 πr3

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

AD = 4r

1129_face centred unit cell.JPG 

 

 

But from the right angled triangle ACD 

AD = √AC2 + DC2 = √a2 + a2= √2a 

√2a = 4r or a = 4/√2 r

∴ volume of cube = (4/√2 r)3 = 64 / 2√2 r3  

percentage of space occupied by sphere 

= volume of sphere / volume of cube × 100 = 16/3 πr3 / 64 /2√2 r3  × 100 = 74%

In body centred cubic unit cell

          1457_body centred cubic unit cell.JPG

Let ‘r’ be the radius of sphere and ‘a’ be the edge length of the cube 

As the sphere at the centre touches the sphere at the corner. Therefore body diagonal 

AD = 4r 

Face diagonal AC = √AB2 + BC2 = √a2 + a2 = √2a 

In right angled triangle ACD = 

AD = √AC2 + CD2 = √2a2 + a2 = √3a 

 √3a = 4r

a = 4r  / √3

∴ Volume of the unit cell = a3 = (4r / √3)3 = 64r3 / 3√3 

No. of spheres in bcc = 2 

∴ volume of 2 spheres = 2 × 4/3πr3 

∴ percentage of space occupied by spheres 

= volume of sphere / volume of cube × 100 = 8/3 πr3 × 100 / 64r3 / 3√3 = 8/3 × 22/7 × 3√3/64 = 68% 

Illustration 10. Show by simple calculation that the percentage of space occupied by spheres in hexagonal cubic packing (hcp) is 74%

 

                              1252_hexagonal cubic packing.JPG

Solution: Let the edge of hexagonal base =a 

And the height of hexagon = h 

And radius of sphere = r 

The centre sphere of the first layer lies exactly over the void of 2nd layer B. 

The centre sphere and the spheres of 2nd layer B are in touch 

So, In Δ PQR (an equilateral triangle) 

PR = 2r, Draw QS tangent at points 

∴ In Δ QRS       ∠QRS = 30°,     SR = r 

Cos30° = SR/QR 

QR = r / √3/2 = 2r / √3  

∴ PQ = √PR– QR2 = √4r2 – 4r2 / 3

h1 = √8r2 / 3 = 2 √2/3 r 

∴ h = 2h1 = 4 √2/3 r 

Now, volume of hexagon = area of base x height 

= 6 × √3 / 4 a2 × h => 6 × √3/4 (2r)2 × 4 √2/3 r 

[∴ Area of hexagonal can be divided into six equilateral triangle with side 2r) 

No. of sphere in hcp = 12 × 1/6 + 1/2 × 2 + 3 

                             = 2+1+3 = 6

∴ Volume of spheres = 6 × 4/3πr3

∴ Percentage of space occupied by sphere = 6 × 4/3 πr3 / 6 × √3/4 × 4r2 × 4√2/3 r × 100 = 74% 

Illustration 11. An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom in each corner of the cube and two atoms on one of its face diagonals. If the volume of this unit cell is 24 x 10-24cm3 and density of the element is 7.20gm/cm3, calculate no. of atoms present in 200gm of the element. 

Solution:Number of atoms contributed in one unit cell

             = one atom from the eight corners + one atom from the two face diagonals 

             = 1+1 = 2 atoms 

             Mass of one unit vell = volume × its density 

             = 24 × 10–24 cm3 × 7.2 gm cm3

             = 172.8 × 10–24 gm 

            ∴ 172.8 10–24 gm is the mass of one – unit cell i.e., 2 atoms

           ∴ 200 gm is the mass = 2 × 200 / 172.8 × 10–24 atoms = 2.3148 × 1024 atoms 

Illustration 12. Calculate the void fraction for the structure formed by A and B atoms such that A form hexagonal closed packed structure and B occupies 2/3 of octahedral voids. Assuming that B atoms exactly fitting into octahedral voids in the HCP formed 
by A

Solution:          Total volume of A atom = 6 × 4 / 3 πrA3

                        Total volume of B atoms = 4 × 4/3 πrA3  4 × 4/3 π(0.414rA)3

                        Since rB/rA as B is in octahedral void of A

                        Volume of HCP = 24√2rA3 

                        Packing fraction = 6 × 4/3 πrA3  + 4 × 4/3 π (0.414rA)3 / 24√2rA3 = 0.7756 

                        Void fraction = 1-0.7756 = 0.2244           

Illustration 13. Lithium iodide crystal has a face-centred cubic unit cell. If the edge length of the unit cell is 620 pm, determine the ionic radius of I- ion. 

Solution: As LiI has fcc arrangement, I- ions will occupy the corners and face-centres. These ions will touch each other along the face diagonal.

                                           1865_face diagonal.JPG 

                        ∴   Face diagonal AB = 4r1 = √a+ a2 = √2a 

                        ∴  r1 = √2a / 4 = 1.414 × 620 pm / 4 = 219.17 pm 

Illustration 14. When heated above 916°C, iron changes its crystal structure from bcc to ccp structure without any change in the radius of atom. The ratio of density of the crystal before heating and after heating is: 

                        (A) 1.089                                              (B) 0.918 

                        (C) 0.725                                              (D) 1.231 

Solution:(B) dbcc / dccp = packing efficiency of bcc / packing efficiency of ccp = 67.92/74.02 = 0.918 

Illustration 15. Select the correct statements:-

                     (A) For CsCl unit cell (edge-length = a), rc + ra = √3/2 a 

For NaCl unit cell (edge-length =), rc + ra = l/2 

The void space in a b.c.c. unit cell is 0.68 

The void space % in a face-centered unit cell is 26% 

Solution:          (A), (B), (C), (D). In bcc structure are r+ + r = √3/2 a 

                        Hence, for CsCl, rC + ra = √3/2 a 

                        ∴ (A) is correct 

                        Since, NaCl crystalise in fcc structure

                        ∴ 2rC + 2ra = edge length of the unit cell 

                        Hence, rC + ra = l/2 

                        ∴ (B) is correct 

                        Since packing fraction of a bcc unit cell is 0.68   

                        ∴ void space = 1–0.68 = 0.32 

                        ∴ (C) is incorrect                                  

                        In fcc unit cell PF = 74% 

                        ∴ VF = 100–74 = 26% 

                        ∴ (D) is correct                                          

Exercise 4. 

(a)What is the empty fraction in case of S.C., B.C.C and F.C.C? 

(b)Which is most packed structure S.C., B.C.C or F.C.C?

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