The atomic radius of an ion which crystallizes in fcc structure is 9/7 – InterviewSolution

1.	The atomic radius of an ion which crystallizes in fcc structure is 9/7 Å. Calculate the lattice constant.
Answer» SOLUTION :Atomic radius, `r = 9/7 Å =9/7 xx 10^(-10)` m. For FCC STRUCTURE, `r = a//(2sqrt2)` or lattice constant, = `9/7 xx 10^(-10) m xx 2 xx 1.4142 = 3.6 xx 10^(-10)` m.

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