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

The unit cell edge length of the alloy is 0.3112 nm

Step-by-step 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.) = (100)/((C_(Cr))/(\rho_(Cr))+ (C_(Ta))/(\rho_(Ta))) \\\\\rho _(Avg.) = (100)/((22)/(7.19)+ (78)/(16.6)) = 12.89\ g/cm^3`

Step 2: determine the average atomic weight;

`A _(Avg.) = (100)/((C_(Cr))/(A_(Cr))+ (C_(Ta))/(A_(Ta))) \\\\A _(Avg.) = (100)/((22)/(52)+ (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 = (nA_(Avg.))/(\rho _(Avg.) N_A) = ((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)^{(1)/(3)}`

Where;

a is the edge length

`a = (3.016*10^(-23))^{(1)/(3)} = (30.16*10^(-24))^{(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