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 mm^3 (1 cm^3) were somehow removed and linked end to end.

Required:
a. How far (in miles) would this chain extend?
b. Now suppose that the density is increased to 1010 mm^-2 by cold working. What would be the chain length of dislocations in 1000 mm^3 of material?

Respuesta :

Answer:

[tex]62.14\ \text{miles}[/tex]

[tex]6213727.37\ \text{miles}[/tex]

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = [tex]10^5\ \text{mm}^{-2}[/tex]

Volume of the metal = [tex]1000\ \text{mm}^3[/tex]

[tex]10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}[/tex]

[tex]1\ \text{mile}=1609.34\ \text{m}[/tex]

[tex]\dfrac{10^5}{1609.34}=62.14\ \text{miles}[/tex]

The chain would extend [tex]62.14\ \text{miles}[/tex]

Dislocation density = [tex]10^{10}\ \text{mm}^{-2}[/tex]

Volume of the metal = [tex]1000\ \text{mm}^3[/tex]

[tex]10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}[/tex]

[tex]\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}[/tex]

The chain would extend [tex]6213727.37\ \text{miles}[/tex]

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