Since r B/r A as B is in octahedral void of A Total volume of A atom = 6 × 4 / 3 πr A 3 Assuming that B atoms exactly fitting into octahedral voids in the HCP formed 2:Ĭalculate the void fraction for the structure formed by A and B atoms such that A form hexagonal closed packed structure and B occupies 2/3 of octahedral voids. ∴ 172.8 10 –24 gm is the mass of one – unit cell i.e., 2 atoms Mass of one unit cell = volume × its density Number of atoms contributed in one unit cell = one atom from the eight corners + one atom from the two face diagonals = 1+1 = 2 atoms of atoms present in 200gm of the element. If the volume of this unit cell is 24 x 10 -24 cm 3 and density of the element is 7.20gm/cm 3, calculate no. Thus, it is concluded that ccp and hcp structures have maximum packing efficiency.Īn element crystallizes into a structure which may be described by a cubic type of unit cell having one atom in each corner of the cube and two atoms on one of its face diagonals. Concepts of Physics by HC Verma for JEE.IIT JEE Coaching For Foundation Classes.
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