Question

Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical density for Nb.

Answer 1

Answer:

The theoretical density for Niobium is $1.87 g/cm^3$.

Explanation:

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density  of the unit cell

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

We have :

Z = 2 (BCC)

M = 92.91 g/mol ( Niobium)

Atomic radius for niobium = r = 0.143 nm

Edge length of the unit cell = a

r = 0.866 a (BCC unit cell)

$a=\frac{0.143 nm}{0.866}=0.165 nm=0.165 \times 10^{-7} cm$

$1 nm = 10^{-7} cm$

On substituting all the given values , we will get the value of 'a'.

$\rho=\frac{2\times 92.91}{6.022\times 10^{23} mol^{-1}\times (0.165 \times 10^{-7} cm)^{3}}$

$\rho =1.87 g/cm^3$

The theoretical density for Niobium is $1.87 g/cm^3$.

Answer 2

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