To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105 mm-2. Suppose that all the dislocations in 1000 mm3 (1 cm3) were somehow removed and linked end to end. How far (in miles) would this chain extend? Now suppose that the density is increased to 109 mm-2 by cold working. What would be the chain length of dislocations in 1000 mm3 of the material?

LD₁ = 10⁵ mm⁻²

LD₂ = 10⁴mm⁻²

V = 1000 mm³

Distance = (LD) (V)

Distance₁ = (10⁵mm⁻²) (1000mm³) = 10*10⁷mm = 10*10⁴m

Distance₂ = (10⁹mm⁻²) (1000mm³) = 1*10¹² mm = 1*10⁹ m

Conversion to miles:

Distance₁ = 10*10⁴ m / 1609m = 62 miles

Distance₂ = 10*10⁹m / 1609 m = 621,504 miles.

A) At Rd₁ = 10^ (5) mm^ (-2), chain length = 62.14 miles

B) At Rd₂ = 10^ (9) mm^ (-2), chain length = 621,372.74 miles

Explanation:

From the question, let's call dislocation density Rd and thus

Rd₁ = 10^ (5) mm^ (-2)

Rd₂ = 10^ (9) mm^ (-2)

And

Volume (V) = 1000 mm³

Now, the formula for chain length dislocation is given as;

L = Rd x V

Thus;

At, Rd = Rd₁ = 10^ (5) mm^ (-2), we have;

L1 = (10^ (5) mm^ (-2)) x (1000mm³) = 10*10^ (7) mm = 100000

At Rd = Rd₂ = 10^ (9) mm^ (-2), we have;

L2 = (10^ (9) mm^ (-2)) (1000mm³) = 10^ (12) mm = 10^ (9) m

Now, let's convert both lengths from metre to miles:

1 mile = 1609.34m

Thus;

L1 = 100000 / 1609. 34m = 62. 14 miles

Similarly,

L2 = 10^ (9) m / 1609.34 m = 621,372.74 miles.