Correct Answer - Option 4 : 0.68, 0.74
Explanation:
The atomic packing factor is defined as the ratio of the volume occupied by the average number of atoms in a unit cell to the volume of the unit cell.
Mathematically, Atomic Packing Factor (APF):
APF \( = \frac{{{N_{atoms}} ~\times ~{V_{atoms}}}}{{{V_{unit\;cell}}}}\) ...(1)
Characteristics of various types of structures are shown in the table below:
|
Characteristics
|
BCC
|
FCC
|
HCP
|
|
a to r relation
|
\(a = \frac{{4r}}{{√ 3 }}\)
|
\(a = 2√ 2 r\)
|
\(a = 2r\)
|
|
The average number of atoms
|
2
|
4
|
6
|
|
Co-ordination number
|
8
|
12
|
12
|
|
APF
|
0.68
|
0.74
|
0.74
|
|
Examples
|
Na, K, V, Mo, Ta, W
|
Ca, Ni, Cu, Ag, Pt, Au, Pb, Al
|
Be, Mg, Zn, Cd, Te
|
For Cubic Unit Cell
Nav = \(N_c\over 8\) + \(N_f\over 2\) + \(N_i \over 1\)
Nav = Average no of atoms in unit cell, Nc = No of corner atoms, Ni = No of interior atoms, Nf = No of face centre atoms
Calculation:
No of atoms in f.c.c unit cell = 4
\(APF = \frac{{{N_{atoms}}{V_{atom}}}}{{{V_{crystal}}}} = \frac{{4\left( {\frac{4}{3}} \right)\pi {r^3}}}{{{{\left( {a } \right)}^3}}}= \frac{{4\left( {\frac{4}{3}} \right)\pi {r^3}}}{{{{\left( {{{2√2r}}{}} \right)}^3}}}\)
for FCC a = 2√2 r where a is side of the cube and r is atomic radius.
APF = 0.74
For BCC:
Nav = \(8\over 8\) + 0 + \(1\over 1\) = 2
√3a = 4r
Put all values in equation 1
(APF)BCC = 0.68
