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An atom crystallises in hexogonal closed pack arrangement. Determine dimensions (radius and length) of a large cyclindrical atom that can be accommodated in the centre of HCP, in terms of radius of host atom. |
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Answer» Solution : The cylinder will pass through centre of middle layer and will lie between the face CENTRES. THEREFORE, Height of cylinder (H) = height of hexagon (h) = 2r Since ,in HCP:H=`4=sqrt((2)/(3))R`, Where r= radius of atoms. `implies h=(4sqrt((2)/(3)-2)` r=1.266 r Also, if R is the radius of cylinder, then in the case of closet contact : `(R)/(r)=0.155implies R=0.155 r,implies h=1.266 r` 
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