Question

Calculate the radius of a vanadium atom, given that it has a BCC crystal structure, density of 5.96 g/cm3, and an atomic weight of 50.9 g/mol (Max. pts. 5).

Answer 1

Answer:

The answer is "$\bold{1.32 \times 10^{-3} \ cm}$".

Explanation:

All of the atoms in a BCC crystalline structure are contained in the 8-corner unit cell.  

Each corner connects the atom to a single cell $\frac{1}{8}$  

Therefore, the unit cell number of atoms:  

$= 8 \times \frac{1}{8}+ 1 \\\\= 1+1 \\\\= 2 \ atom$

$The mass unit cell = \frac{ \text{Number of atoms} \times \text{atomic weight}} {Avagadro number}\\\\= \frac{2 \times 50.9}{6.023 \times 10^{23}} \\\\= 1.69 \times 10^{-22} \ g\\\\Area Of the atom= \frac{4r}{\sqrt{3}}\\\\ 5.96 = \frac{1.69 \times 10^{-22}}{volume}\\\\volume= 2.835 \times 10^{25}\\\\v=d^3\\\\v= (\frac{4r}{\sqrt{3}})^3\\\\\to 2.835 \times 10^{-23} \times (\sqrt{3})^3 = 4^3 r^3$  

$\to \sqrt[3]{\frac{{2.835 \times 10^{-23} \times (\sqrt{3})^3}}{4^3}} =r\\\\\to r= 1.32 \times 10^{-3} \ cm$