Solveeit Logo

Question

Question: Find packing fraction of three dimensional unit cell of AAAAA……………type hypothetical arrangement in w...

Find packing fraction of three dimensional unit cell of AAAAA……………type hypothetical arrangement in which hexagonal packing is taken in layer.

Explanation

Solution

To solve this we must first calculate the total area occupied by the spheres in the unit cell and the total area of the square unit cell. The packing efficiency can then be calculated by the ratio of area of spheres in the unit cell to the total area of the square unit cell.

Complete step-by-step answer:
The structure of the hcp unit cell is as follows:

From the structure, we can see that
a=2ra = 2r …… (1)
Where aa is the side of the face of the hexagon,
rr is the radius of the sphere.
We know the formula for the volume of the hexagon,
V=332a2×hV = \dfrac{{3\sqrt 3 }}{2}{a^2} \times h
Where VV is the volume of the hexagon,
aa is the side of the face of the hexagon,
hh is the height of the hexagon and is equal to 223a2\sqrt {\dfrac{2}{3}} a.
Thus, volume of hexagon is,
V=332a2×223aV = \dfrac{{3\sqrt 3 }}{2}{a^2} \times 2\sqrt {\dfrac{2}{3}} a
From equation (1), substitute a=2ra = 2r. Thus,
V=332(2r)2×2232rV = \dfrac{{3\sqrt 3 }}{2}{\left( {2r} \right)^2} \times 2\sqrt {\dfrac{2}{3}} 2r
V=242r3V = 24\sqrt 2 {r^3} …… (2)
Where VV is the volume of the hexagon,
rr is the radius of the sphere.
We know the formula for the volume of the sphere is,
V=43πr3V = \dfrac{4}{3}\pi {r^3}
Where VV is the volume of the sphere,
rr is the radius of the sphere.
In a hcp structure, there are a total six spheres. Thus, total volume of the spheres is,
V=6(43πr3)V = 6\left( {\dfrac{4}{3}\pi {r^3}} \right) …… (3)
The packing efficiency is the ratio of volume of spheres in the unit cell to the total volume of the cube cell. Thus, to calculate the packing efficiency divide equation (3) by equation (2). Thus,
Packing efficiency=6(43πr3)242r3{\text{Packing efficiency}} = \dfrac{{6\left( {\dfrac{4}{3}\pi {r^3}} \right)}}{{24\sqrt 2 {r^3}}}
Packing efficiency=π32{\text{Packing efficiency}} = \dfrac{\pi }{{3\sqrt 2 }}
Substitute π=3.14\pi = 3.14. Thus,
Packing efficiency=3.1432{\text{Packing efficiency}} = \dfrac{{3.14}}{{3\sqrt 2 }}
Packing efficiency=0.7402{\text{Packing efficiency}} = 0.7402
Thus, the percentage packing efficiency is 0.7402×100%=74.02%0.7402 \times 100\% = 74.02\% .

Thus, packaging fraction of three dimensional unit cell of AAAAA……………type hypothetical arrangement in which hexagonal packing is taken in layer is 74%74\% .

Note: In hcp structure, each corner atom is shared by six unit cells and thus, the contribution of the corner atoms is two atoms. The tom at the centre is shared by two unit cells and thus, the contribution of the centre atom is one. One single cell contains three atoms. Thus, the total number of atoms in a hcp unit cell are six.