Let the height of hexagon be ‘c’; edge length be ‘a’, and radius of sphere be ‘r’
Volume of unit cell = Area of base x height
Area of base = 6 x Area of equilateral triangle 123
From the figure; a = 2r
• 4 is placed over the triangular void in Layer B
• When a perpendicular from the centre of 4 is dropped on Layer A; it coincides with the centroid (G) of triangle 123. The length of the perpendicular is c/2.
• Centre of sphere 4, centroid of triangle 123 and centre of sphere 3 form a right angled triangle shown as triangle G43 above
• If we can find the value of x, value of c /2 can be found out from triangle G43
• In triangle 123, M3 is the median and G is the centroid.
• Triangle M13 is a right angled triangle. So
• G divides M3 in the ratio of 1:2, so the value of G3 = x
• Now, in triangle G43:
• Substitute a = 2r in above equation we get
= 0.74
% Packing Fraction = 74%
