Prove mathematically that the empty space in a simple cubic cell is 47.6%. The volume of a sphere is 4/3*3.14r^3.

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In a simple cubic lattice, there is a sphere loacted at each corner of the latice. Only 1/8 of the area of each sphere is inside the unit cell. However, there are 8 spheres (corners). So the total area occupied in the cell is 1/8 * 8 * Area of a sphere.

For your problem, we'll assume a tightly packed cell, so that the radius of the spheres is r and each side of the cube is 2r long.

% empty space = 100% - area of spheres in cube / area of cube

= 100% - 4/3*3.14r^3 / (2r)^3

= 100% - 4/24*3.14

= 100% - 52.3%

= 47.6%

Let us assume that the cubic cell has each of its side meauring 2a units. Then the volume of this cube = (2a)^3 = 8a^3 cubic units.

The sphere inside this cube touching all the six faces should have a diameter of 2a , or radius = a. So the volume of the sphere of radius a = (4/3)(pi)a^3 cubic units, where pi = 3.14592654, the circle's constant ratio of circumference to diameter.

Therefore, empty space in the cube of side 2a when occupied by the sphere of radius a = volume of the cube - volume of the sphere = 8a^3 - (4/3)(3.141592654)(a^3) = (8-4.188790205) a^3 = 3.811209795a^3 cubic units.

Therefore the percentage of empty space = (Volume of mpty space/Total volume of the cube)*100

= [(3.811209795a^3)/8a^3)]*100

= 47.64012244 %

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