To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of \(10^{4} \mathrm{mm}^{-2}\). Suppose that all the dislocations in \(1000 \mathrm{~mm}^{3}\left(1 \mathrm{~cm}^{3}\right)\) were somehow removed and linked end to end. How far (in miles) would this chain extend? Now suppose that the density is increased to \(10^{10} \mathrm{~mm}^{-2}\) by cold working. What would be the chain length of dislocations in \(1000 \mathrm{~mm}^{3}\) of material?

First, we need to calculate the total number of dislocations in the specimen for both densities, which can be obtained by multiplying the density with the volume of the specimen. For the first case with a dislocation density of \(10^4 \,\mathrm{mm}^{-2}\), the total number of dislocations is: Total dislocations = Density × Volume Total dislocations = \(10^4 \,\mathrm{mm}^{-2} \times 1000 \,\mathrm{mm}^{3}\) Total dislocations = \(10^7\) dislocations For the second case with a dislocation density of \(10^{10} \,\mathrm{mm}^{-2}\), the total number of dislocations is: Total dislocations = \(10^{10} \,\mathrm{mm}^{-2} \times 1000 \,\mathrm{mm}^{3}\) Total dislocations = \(10^{13}\) dislocations

Next, we need to find the chain lengths of the dislocations for both cases. To do this, we assume that each dislocation has a length equal to the average distance between them, which can be estimated as the inverse of the square root of the dislocation density. For the first case with a dislocation density of \(10^4 \,\mathrm{mm}^{-2}\): Average dislocation distance = \(\sqrt{1/(10^4 \,\mathrm{mm}^{-2})}\) Chain length = Total dislocations × Average dislocation distance Chain length = \(10^7 \times \sqrt{1/(10^4 \,\mathrm{mm}^{-2})}\) Chain length = \(10^7 \times \sqrt{1/10^4} \,\mathrm{mm}\) Chain length = \(10^7 \times 0.01 \,\mathrm{mm} = 10^5 \,\mathrm{mm}\) For the second case with a dislocation density of \(10^{10} \,\mathrm{mm}^{-2}\): Average dislocation distance = \(\sqrt{1/(10^{10} \,\mathrm{mm}^{-2})}\) Chain length = Total dislocations × Average dislocation distance Chain length = \(10^{13} \times \sqrt{1/(10^{10} \,\mathrm{mm}^{-2})}\) Chain length = \(10^{13} \times \sqrt{1/10^{10}} \,\mathrm{mm}\) Chain length = \(10^{13} \times 0.0001 \,\mathrm{mm} = 10^9 \,\mathrm{mm}\)

Finally, we need to convert the chain lengths from millimeters to miles. We know that 1 mile = 1,609,344 millimeters. For the first case with a dislocation density of \(10^4 \,\mathrm{mm}^{-2}\): Chain length (in miles) = Chain length (in mm) / 1,609,344 mm/mile Chain length (in miles) = \(10^5 \,\mathrm{mm} / 1,609,344 \,\mathrm{mm/mile}\) Chain length (in miles) ≈ 0.062 miles For the second case with a dislocation density of \(10^{10} \,\mathrm{mm}^{-2}\): Chain length (in miles) = Chain length (in mm) / 1,609,344 mm/mile Chain length (in miles) = \(10^9 \,\mathrm{mm} / 1,609,344 \,\mathrm{mm/mile}\) Chain length (in miles) ≈ 621 miles In conclusion, for the first case, the chain of dislocations would extend about 0.062 miles, and for the second case, it would extend about 621 miles.