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Calculate 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. Assuming that B atoms exactly fitting into octahedral voids in the HCP formed by A

1 Answer

Total volume of A atom =$\large\frac{6\times 4}{3}$$\pi r_A^3$
Total volume of B atoms =$4\times \large\frac{ 4}{3}$$\pi r_B^3=4\times \large\frac{ 4}{3}$$\pi (0.414r_A^3)$
Since $\large\frac{r_A}{r_B}$$=0.414$ as B is in octahedral void of A
Volume of HCP = 24$\sqrt 2r_A^3$
Packing fraction =$ \large\frac{\Large\frac{6\times 4}{3}\pi r_A^3+4\times \Large\frac{ 4}{3}\pi (0.414r_A^3)}{24\sqrt2r_A^3}$
Void fraction = 1-0.7756 = 0.2244
Hence (A) is the correct answer.
answered May 30, 2014 by sreemathi.v

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