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

The answer is "
`\bold{1.32 * 10^(-3) \ cm}`".

Step-by-step 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
`(1)/(8)`

Therefore, the unit cell number of atoms:

`= 8 * (1)/(8)+ 1 \\\\= 1+1 \\\\= 2 \ atom`

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

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