Chemistry-
General
Easy

Question

In simple cubic lattice, the spheres are packed in the form of a square array by laying down a base of spheres and then piling upon the base other layers in such a way that each sphere is immediately above the other sphere. In this structure, each sphere is in contact with six nearest neighbours. The percentage of occupied volume in this structure can be calculated as follows
The edge length ‘a’ of the cube will be twice the radius of the sphere, i e comma blank a equals 2 r. Since, in the primitive cubic lattice, there is only one sphere present in the unit lattice, the volume occupied by the sphere is

V equals fraction numerator 4 over denominator 3 end fraction pi r to the power of 3 end exponent or V equals fraction numerator 4 over denominator 3 end fraction pi open parentheses fraction numerator a over denominator 2 end fraction close parentheses to the power of 3 end exponent
The fraction of the total volume occupied by the sphere is
ϕ equals fraction numerator fraction numerator 4 over denominator 3 end fraction pi open parentheses fraction numerator a over denominator 2 end fraction close parentheses to the power of 3 end exponent over denominator a to the power of 3 end exponent end fraction equals fraction numerator pi over denominator 6 end fraction equals 0.5236 blank o r blank 52.36 percent sign
In a simple cubic cell, an atom at the corner contributes to the unit cell

  1. fraction numerator 1 over denominator 4 end fraction part    
  2. fraction numerator 1 over denominator 2 end fraction part    
  3. 1 part    
  4. fraction numerator 1 over denominator 8 end fraction part    

The correct answer is: fraction numerator 1 over denominator 8 end fraction part


    An atom on the corner contributes to the unit cell equals fraction numerator 1 over denominator 8 end fraction part

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