Respuesta :
Answer:
The theoretical density for Niobium is [tex]1.87 g/cm^3[/tex].
Explanation:
Formula used :
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density of the unit cell
Z = number of atom in unit cell
M = atomic mass
[tex](N_{A})[/tex] = 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)
[tex]a=\frac{0.143 nm}{0.866}=0.165 nm=0.165 \times 10^{-7} cm[/tex]
[tex]1 nm = 10^{-7} cm[/tex]
On substituting all the given values , we will get the value of 'a'.
[tex]\rho=\frac{2\times 92.91}{6.022\times 10^{23} mol^{-1}\times (0.165 \times 10^{-7} cm)^{3}}[/tex]
[tex]\rho =1.87 g/cm^3[/tex]
The theoretical density for Niobium is [tex]1.87 g/cm^3[/tex].
