A) 288 pm
B) 408 pm
C) 144 pm
D) 204 pm
Correct Answer: A
Solution :
For fee lattice, \[4r=\sqrt{2}a\] \[r=\frac{\sqrt{2}}{4}a=\frac{a}{2\sqrt{2}}\] \[=\frac{408}{2\sqrt{2}}\] \[=144\,pm\] diameter d = 2r = 2 x 144 pm = 288pmYou need to login to perform this action.
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