0.284
Iron Has BCC Crystal Structure With Lattice Parameter Of 0.142 :calculate Atomic Diameter?
The Lattice constant is a=(2*atomic radius)/sqrt(3)
if a=0.142 then radius = (sqrt(3)*a)/2
or r = (1.73*0.142)/2
= 0.123 nm
so the Diameter is 2r or 0.246 nm
if a=0.142 then radius = (sqrt(3)*a)/2
or r = (1.73*0.142)/2
= 0.123 nm
so the Diameter is 2r or 0.246 nm
