InterviewSolution
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InterviewSolution
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A hexagonal close-packed lattic can be represented by figures (a) and (b) below. If `c=asqrt((8)/(3))=1.633a` , there is an atom at each corner of the unit cell and another atom which can be located by moving one-third the distance along the diagonal of the rhombus base, starting at the lower left hand corner and moving perpendicularly upward by c/2 . Mg crystallizes in this lattice and has a density of `1.74 g cm^(-3)`. What is the volume of unit cell?A. 46B. 20C. 180D. 90 |
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Answer» Correct Answer - A `Z_(eff)`=8 (corners)`xx(1)/(8)` (per corner share)+1 (in the body)=2 atoms/unit cell. Aw of Mg=24.3g Mass=(2 atoms) `((24.3g)/(6.02xx10^(23)"atoms"))=8.07xx10^(-23)g` volume= `("Mass")/("Density")=((8.07xx10^(-23)g)/(1.74gcm^(-3)))xx(10^(8)overset(@)A)^(3)=46.4overset(@)A^(3)` |
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