Question

An aluminum crystal is bent into a radius of curvature of 5 cm. The minimum dislocation density in the material is (Burgers vector = \(3 \, \text{\AA}\)):

• A. \(p = 3.42 \times 10^{14} \, \text{m}^{-2}\)
• B. \(p = 3.42 \times 10^{15} \, \text{m}^{-2}\)
• C. \(p = 3.42 \times 10^{16} \, \text{m}^{-2}\)
• D. \(p = 3.42 \times 10^{17} \, \text{m}^{-2}\)

Hint:

Dislocation density quantifies the number of defects in a material. The radius of curvature R inversely affects dislocation density, making it crucial in bending-related calculations.

Answer 1

The Correct Option is A

\(\text{Dislocation density is calculated as:} \)
\(p = \frac{1}{bR}\)
\(\text{where } b = 3 \text{ \AA} = 3 \times 10^{-10} \text{ m and } R = 5 \text{ cm} = 5 \times 10^{-2} \text{ m:}\)
\(p = \frac{1}{(3 \times 10^{-10})(5 \times 10^{-2})} = 3.42 \times 10^{14} \text{ m}^{-2}\)