A) \[1.27\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[5.08\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[2.54\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[3.59\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: D
Solution :
Interatomic spacing for a fee lattice \[r={{\left[ {{\left( \frac{a}{2} \right)}^{2}}+{{\left( \frac{a}{2} \right)}^{2}}+{{(0)}^{2}} \right]}^{1/2}}=\frac{a}{\sqrt{2}}\] \[a\] being lattice constant. \[\therefore \] \[a=\sqrt{2}r=\sqrt{2}\times 2.54=3.59\,\overset{\text{o}}{\mathop{\text{A}}}\,\] NOTE: Interatomic spacing is just the nearest neighbors distance.
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