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

For CsCl type structures, 2(r

+

+r

−

)=a

3

So, a=

3

2

(0.165+0.181)=0.4 mm

Density =

N

A

×a

3

Z×M

=

6.02×10

23

×(4×10

−10

)

3

1×(133+35.5)

=4.37× 10

3

kg m

3

(by converting units)

Answer 2

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)

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