Mumbai University > First Year Engineering > Sem 1 > Applied Physics 1

Marks: 7M

Year: May 2016

Given:

Nearest neighbor distance, $a = 2r=0.27 nm=0.27 \times 10^{-9} m$

Atomic weight of zinc=65.37

The height of the unit cell, h= 0.494 nm

Find:

(i) Volume of unit cell

(ii) Density

(iii) Atomic Packing fraction

Solution:

1. The volume of the unit cell, V:

   The volume of the HCP unit cell is given by:

   $V = \frac{3\sqrt{3}a^2c}{2}$

   Substituting the values, we get

   $V = \frac{3\sqrt{3}(0.27 \times 10^{-9})^2 \times 0.494\times 10^{-9}}{2}$

   $V = 9.35 \times 10^{-29}m^3$

2. The density of zinc:

   Effective number of atoms in unit cell of HCP, n=6

   using the relation $\rho = \frac{nM}{N_A \times V}$, we have

   $[where \hspace{0.3cm}n = 6, M = 65.37 (given), N_A = 6.0238 \times 10^{36}]$

   $\rho = \frac{6 \times 65.37}{6.0238 \times 10^{26}\times 9.35 \times 10^{-29}} \approx 6964kg/m^3$

3. Atomic Packing Factor:

   Now, c=1.633 a

   $V_{unit cell}=1.633(2r)^3sin 60^0$

   $V_{atom} = \frac{4}{3}πr^3$

   $APF=\frac{2(\frac{4}{3} πr^3)}{(1.633(2r)^3 sin60^0}=0.741$