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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?

User Rogerio
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1 Answer

3 votes

Answer:


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}

User Cpres
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