Question

Cromium and tantalum both have the BCC crystal structure, and Ta forms a substitutional solid solution for all concentrations at room temperature. Compute the unit cell edge length for a 22 wt% Cr - 78 wt% Ta alloy. The room-temperature density and atomic weight of Cr are 7.19 g/cm3 and 52.00 g/mol, the room-temperature density and atomic weight of Ta are 16.6 g/cm3 and 180.95 g/mol, respectively.

Answer 1

Answer:

The unit cell edge length of the alloy is 0.3112 nm

Explanation:

Given;

density of chromium, Cr= 7.19 g/cm³

atomic weight of chromium, Cr = 52.00 g/mol

density of tantalum, Ta = 16.6 g/cm³

atomic weight of tantalum, Ta  = 180.95 g/mol

Step 1: determine the average density of the atoms;

$\rho _{Avg.} = \frac{100}{\frac{C_{Cr}}{\rho_{Cr}}+ \frac{C_{Ta}}{\rho_{Ta}}} \\\\\rho _{Avg.} = \frac{100}{\frac{22}{7.19}+ \frac{78}{16.6}} = 12.89\ g/cm^3$

Step 2: determine the average atomic weight;

$A _{Avg.} = \frac{100}{\frac{C_{Cr}}{A_{Cr}}+ \frac{C_{Ta}}{A_{Ta}}} \\\\A _{Avg.} = \frac{100}{\frac{22}{52}+ \frac{78}{180.95}} = 117.08 \ g/mol$

Step 3: determine the cubic volume of the atoms;

there are 2 atoms per unit cell of a body centered cubic structure.

$V_C = \frac{nA_{Avg.}}{\rho _{Avg.} N_A} = \frac{(2)*117.08}{12.89(6.023*10^{23})}\\\\V_C =3.016 *10^{-23} \ cm^3/unit \ cell$

Step 4: determine the unit cell edge length;

Vc = a³

$a = (V_C)^{\frac{1}{3}}$

Where;

a is the edge length

$a = (3.016*10^{-23})^{\frac{1}{3}} = (30.16*10^{-24})^{\frac{1}{3}} = 3.112*10^{-8} \ cm$

$a = 3.112 *10^{-10} \ m =0.3112 *10^{-9} \ m = 0.3112\ nm$

Therefore, the unit cell edge length of the alloy is 0.3112 nm