Question

Determine the lattice constant for CsCl ,given the atomic radius of Cs and Cl as 0.71 and 0.181 nm respectively

Answer 1

Explanation:

For CsCl type structures, 2(r++r−)=a3

So, a= 32(0.165+0.181)=0.4 mm

Density =NA×a3Z×M=6.02×1023×(4×10−10)31×(133+35.5) =4.37× 103 kg m3 (by converting units).