Compute the percentage void space per unit volume of unit cell in zinc-fluoride structure .A. `25.07 %`B. `74.93 %`C. `52%`D. `48%`

Compute the percentage void space per unit volume of unit cell in zinc

1.	Compute the percentage void space per unit volume of unit cell in zinc-fluoride structure .A. `25.07 %`B. `74.93 %`C. `52%`D. `48%`
Answer» Correct Answer - A 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 Q for tetrahedral holes, `=0.225=(1+(0.255)^(3))=0.7493` `therefore` void volume=1-0.7493 =0.2507 / unit volume of unit cell % void space=25.07%

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Compute the percentage void space per unit volume of unit cell in zinc-fluoride structure .A. `25.07 %`B. `74.93 %`C. `52%`D. `48%`

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