Question

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?

Answer 1

`62.14\ \text{miles}`

`6213727.37\ \text{miles}`

Step-by-step explanation:

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

Dislocation density =
`10^5\ \text{mm}^(-2)`

Volume of the metal =
`1000\ \text{mm}^3`

`10^5* 1000=10^8\ \text{mm}\\ =10^5\ \text{m}`

`1\ \text{mile}=1609.34\ \text{m}`

`(10^5)/(1609.34)=62.14\ \text{miles}`

The chain would extend
`62.14\ \text{miles}`

Dislocation density =
`10^(10)\ \text{mm}^(-2)`

Volume of the metal =
`1000\ \text{mm}^3`

`10^(10)* 1000=10^(13)\ \text{mm}\\ =10^(10)\ \text{m}`

`(10^(10))/(1609.34)=6213727.37\ \text{miles}`

The chain would extend
`6213727.37\ \text{miles}`