A) 0.1624 nm
B) 0.1246 nm
C) 0.2164 nm
D) 0.1426 nm
Correct Answer: B
Solution :
For an fee lattice,\[{{d}_{face}}=a\sqrt{2}\]or\[4{{r}_{Ni}}=a\sqrt{2}\] \[{{r}_{Ni}}=\frac{a\sqrt{2}}{4}=\frac{\sqrt{2}}{4}\times 0.3524=0.1246nm\]
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