In a face-centred cubic $(FCC)$ unit cell,what is the total volume occupied by the atoms?
- A$\frac{4}{3} \pi r^3$
- B$\frac{8}{3} \pi r^3$
- C$\frac{16}{3} \pi r^3$
- D$\frac{64 r^3}{3 \sqrt{3}}$
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Similar Questions
$A$ compound is formed by elements $X$, $Y$, and $Z$. Atoms of $Z$ (anions) make $FCC$ lattice. Atoms of $X$ (cations) occupy all the octahedral voids. Atoms of $Y$ (cations) occupy $\frac{1}{3}$ rd of the tetrahedral voids. The formula of the compound is:
Which of the following statements is not correct?
Hexagonal close packing $(HCP)$ is found in the crystal lattice of:
Medium
View SolutionIn a $ccp$ structure of $X$ atoms,$Y$ atoms occupy the octahedral voids. If two $X$ atoms at the corners are replaced by $Z$ atoms,the molecular formula of the solid will be:
Medium
View Solution$A$ compound is formed by cation $C$ and anion $A$. The anions form hexagonal close packed $(hcp)$ lattice and the cations occupy $75 \%$ of octahedral voids. The formula of the compound is
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