A) \[\frac{16}{3}\pi {{\operatorname{r}}^{3}}\]
B) \[\frac{20}{3}\pi {{\operatorname{r}}^{3}}\]
C) \[\frac{24}{3}\pi {{\operatorname{r}}^{3}}\]
D) \[\frac{12}{3}\pi {{\operatorname{r}}^{3}}\]
Correct Answer: A
Solution :
Volume of 1 unit cell \[=\frac{4}{3}\pi {{r}^{3}}\] \[{{\operatorname{X}}_{eff}} in FCC=4\] \[\therefore \]Volume of atom per FCC unit cell \[=4\times \frac{4}{3}\pi {{r}^{3}} =\frac{16}{3}\pi {{r}^{3}}\]
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