Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?
We know, for a fcc unit cell,radius of atom
r = a/2√2
Or
a = 2r√2
By putting the values of r &√2 we get
a = 2 x 0.144 x 1.414 = 0.407nm
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 21: Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a si....
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Can u also find density