Question

Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm−3, calculate atomic radius of niobium using its atomic mass 92.90 u.

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Answer 1

Answer :

• Atomic radius = $\rm{1.43\times{}10^{-10}\:m.}$

Step-by-step explanation :

Given that,

• Density, ρ = $\rm{8.55\:g\:cm^{-3}}$
• Atomic mass, M = $\rm{92.90\:g\:mol^{-1}}$

Also,

• For bbc structure, Z = 2
• Avogadro's number, $\rm{N_0=6.023\times{}10^{23}}$

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We know that, $\rm{\rho=\dfrac{Z\times{}M}{a^3\times{}N_0}}$

$\bold{OR}$

$\rm{a^3=\dfrac{Z\times{}M}{\rho\times{}N_0}}$

Putting the given values, we get,

$\implies\rm{a^3=\dfrac{2\times{}92.90\:g\:mol^{-1}}{8.55\:g\:cm^{-3}\times{}6.023\times{}10^{23}\:mol^{-1}}}$

$\implies\rm{a^3=36.09\times{}10^{-24}\:cm^3}$

$\implies\rm{a=3.3\times{}10^{-8}\:cm}$

$\implies\rm{a=3.3\times{}10^{-10}\:m}$

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Now, we know for bbc structure,

$\implies\rm{r=\dfrac{\sqrt{3}}{4}a}$

$\implies\rm{r=\dfrac{\sqrt{3}}{4}\times{}3.3\times{}10^{-10}\:m}$

$\implies\rm{r=}\bold{1.43\times{}10^{-10}\:m}$

Answer 2

Answer:

Given density (d) = 8.55 g/cm³

Atomic mass (M) = 93u = 93 g/mol

We know, Avogadro number, N = 6.022 × 10²³

given lattice is body - centered cubic,

so, number of atoms per unit cell (z) = 2

we know, \bf{d=\frac{zM}{a^3N}}

=> 8.55 = (2 × 93)/(a³ × 6.023 × 10²³)

=> 8.55 × a³ × 6.023 × 10²³ = 186

=> a³ = 186/(8.55 × 6.023 × 10²³)

=> a³ = 36.124 × 10^-24

=> a = 3.3057 × 10^-8 cm

for BCC unit cell, radius = √3a/4

= (√3 × 3.3057 × 10^-8)/4

= (1.732 × 3.3057)/4 × 10^-8

= 1.431 × 10^-8 cm = 1.431 × 10^-10 m

= 1.431 A°

hence, atomic radius of Niobium is 1.431A°.