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The Solid State

Question
CBSEENCH12005492

Niobium crystallises in body centered cubic structure. If density is 8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 93 u.

Solution

We have give that,
Density (d)=8.55 g Cm-3
Atomic mass (M) =93u =93 gMol-1
Atomic radius (r) = ?

We know that, Avogadro number Na =6.022 x 10 23 mol-1

since given lattice is bcc

therefore
Number of atoms per unit cell (z) =2

we know that

 

straight d equals space fraction numerator zM over denominator straight a cubed straight N subscript straight A end fraction
8.55 space straight g space cm to the power of negative 3 end exponent space equals fraction numerator 2 space straight x space 93 space over denominator straight a cubed space straight x 6.022 space straight x 10 to the power of 23 space mol to the power of negative 1 end exponent end fraction

straight a cubed space equals space fraction numerator 2 space straight x 93 over denominator 8.55 space straight x space 6.022 space straight x space 10 to the power of 23 end fraction cm cubed

straight a cubed space equals 3.6124 space straight x space 10 to the power of negative 23 end exponent

straight a cubed space equals 36.124 space straight x space 10 to the power of negative 24 end exponent
straight a equals space 3.3057 space straight x 10 to the power of negative 8 end exponent space cm

For space bb space unit space cell space radius space left parenthesis straight r right parenthesis space equals fraction numerator square root of 3 over denominator 4 end fraction straight a

straight r equals space fraction numerator 1.732 over denominator 4 end fraction space straight x space 3.3057 space straight x 10 to the power of negative 8 end exponent

straight r equals fraction numerator 5.725 over denominator 4 end fraction space straight x space 10 to the power of negative 8 end exponent cm
straight r equals space 14.31 space straight x space 10 to the power of negative 9 end exponent space cm

straight r equals 14.31 nm