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Compute the percentage void space per unit volume of unit cell in zinc-fluoride structure. |
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Answer» SOLUTION :Since, anions OCCUPY fcc POSITIONS and half of the tetrahedral HOLES are occupied by cations. Since, there are four anions and 8 tetrahedral holes per unit cell, the fraction of volume occupied by spheres/unit volume of the unit cell is `=4 xx (4/3 pi r_(a)^(3) +1/2 xx 8 xx (4/3 pi r_(c )^(3)))/(16sqrt(2)r^(3)) = pi/(3sqrt(2))[1+(t_(c)/t_(a))^(3)]` `therefore` For tetrahedral holes. `r_(0)/r_(a) = 0.225 = pi/(3sqrt(2))(1 + (0.225)^(3)) = 0.7496` `therefore` Void volume =1-0.7496 = 0.2504 / unit volume of unit cell. % Void space = 25.04 % |
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