The atomic weight, density, and atomic radius for three hypothetical alloys are listed in the following table. For each, determine whether its crystal structure is FCC, BCC, or simple cubic and then justifiy your determination. A simple cubic unit cell is shown in Figure 3.24.
| Alloy | Atomic Weight (\(g \over mol\)) | Density (\(g \over {cm}^3\)) | Atomic Radius (nm) |
|---|---|---|---|
| A | 77.4 | 8.22 | 0.125 | B | 107.6 | 13.42 | 0.133 |
| C | 127.3 | 9.23 | 0.142 |
50Cr: 4.34% A = 49.9460 amu
52Cr: 83.79% A = 51.9405 amu
53Cr: 9.50% A = 52.9407 amu
54Cr: 2.37% A = 53.9389 amu
Prove that \(A_{Cr}\) = 51.9963 amu
\(A = f_{1} * A_{1} + f_{2} * A_{2} + f_{3} * A_{3} + f_{4} * A_{4} \)
4.34% * 49.9460 + 83.79% * 51.9405 + 9.50% * 52.9407 + 2.37% * 53.9389 = 51.9963 amu
51.9963 amu = 51.9963 amu \( \implies \) correct