Question

Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution for concentrations up to approximately 20 wt% V at room temperature. Compute the unit cell edge length for a 92 wt% Fe-8 wt% V alloy.

Answer 1

Answer:

The unit cell edge length for the alloy is 0.288 nm

Explanation:

Given;

concentration of vanadium, Cv = 8 wt%

concentration of iron, Cfe = 92 wt%

density of vanadium = 6.11 g/cm³

density of iron = 7.86 g/cm³

atomic weight of vanadium, Av = 50.94 g/mol

atomic weight of iron, Afe= 55.85 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_v}{\rho _v} + \frac{C__{Fe}}{\rho _{Fe}} }$

$\rho _{Avg.} = \frac{100}{\frac{8}{6.11} + \frac{92}{7.86} } = 7.68 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _v} + \frac{C__{Fe}}{A _{Fe}} }$

$A _{Avg.} = \frac{100}{\frac{8}{50.94} + \frac{92}{55.85} } = 55.42 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a BCC crystal structure, there are 2 atoms per unit cell; n = 2

$V_c=\frac{2*55.42}{ 7.68*6.022*10^{23}} = 2.397*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (2.397*10^{-23}}){^\frac{1}{3}}\\\\a = 2.88 *10^{-8} \ cm$= 0.288 nm

Therefore, the unit cell edge length for the alloy is 0.288 nm