A) hcp lattice - A, \[\frac{2}{3}\] Tetrahedral voids - B
B) hcp lattice - A, \[\frac{1}{3}\] Tetrahedral voids - B
C) hcp lattice - B, \[\frac{2}{3}\] Tetrahedral voids - A
D) hcp lattice - B, \[\frac{1}{3}\] Tetrahedral voids - A
Correct Answer: D
Solution :
Here, \[{{\operatorname{A}}_{2}}{{B}_{3}}\] can also be written as \[{{\operatorname{A}}_{4}}{{B}_{6}}\]. Since, hcp has six atoms, so ?B? forms hcp lattice and ?A? is present in void. \[\operatorname{Total} tetrahedral voids = 12\] \[\therefore \] Fraction of tetrahedral voids occupied by \[A=4/12=\frac{1}{3}\]
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