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. (Jr., 01/2018, p. P-5) Jr., W. C., Rethwisch, D. G. (2018). Materials Science and Engineering: An Introduction, Enhanced eText, 10th Edition. [[VitalSource Bookshelf version]]. Retrieved from vbk://9781119405498 Always check citation for accuracy before use.

Answer 1

Answer:

The theoretical density of the Niobium is 8.57 g/cm³.

Explanation:  

For a BCC crystal structure the density can be calculated using the following equation:

$\rho = \frac{nA}{VN}$

Where:

n: is the number of atoms, for a BCC = 2 atoms

A: is the atomic weight = 92.91 g/mol

V: is the unit cell's volume

N: is the Avogadro's number = 6.022x10²³ atom/mol

The volume of the unit cell of a BCC crystal structure (V) can be found as follows:

$V = (\frac{4R}{\sqrt{3}})^{3}$

Where R is the atomic radius = 0.143 nm

$V = ({\frac{4R}{\sqrt{3}}})^{3} = ({\frac{4*0.143 nm*\frac{1 cm}{1\cdot 10^{7} nm}}{\sqrt{3}}})^{3} = 3.60 \cdot 10^{-23} cm^{3}$    

Now, the density is:

$\rho = \frac{nA}{VN} = \frac{2 atom*92.91 g/mol}{3.60 \cdot 10^{-23} cm^{3}*6.022 \cdot 10^{23} atom/mol} = 8.57 g/cm^{3}$

Therefore, the theoretical density of the Niobium is 8.57 g/cm³.

I hope it helps you!