Chapter 7, Problem 1QAP

To determine

The distance of the chain length extended is to be determined and the chain length is to be determined.

Answer to Problem 1QAP

The chain extends up to $0.0646\text{2 mile}$ , and the length of the chain is $0.6275\text{ mile}$ .

Explanation of Solution

Given:

Density of the dislocation, $\rho =10{\text{4 mm}}^{-2}$

Volume of the dislocation, $V=1000{\text{ mm}}^3$ .

The formula used for the length of the atomic chain without dislocation.

  $L_w=\rho V$

Here, $L_w$ is the length of the atomic chain without dislocation.

Substitute $10{\text{4 mm}}^{-2}$ for $\rho$ , and $1000{\text{ mm}}^3$ for $V$ in the above equation.

  $\begin{array}{c}{L}_w=10{\text{4 mm}}^{-2}\times 1000{\text{ mm}}^3\\ =10\text{4}000\left|\text{mm}\right|\times \frac{\text{6}\text{.213}\times {\text{10}}^{-7}\text{ mile}}{1\left|\text{mm}\right|}\\ =0.0646\text{ mile}\end{array}$

For chain length:

Substitute $1010{\text{ mm}}^{-2}$ for $\rho$ , and $1000{\text{ mm}}^3$ for $V$ in the above equation.

  $\begin{array}{c}{L}_w=1010{\text{ mm}}^{-2}\times 1000{\text{ mm}}^3\\ =1010000\left|\text{mm}\right|\times \frac{\text{6}\text{.213}\times {\text{10}}^{-7}\text{ mile}}{1\left|\text{mm}\right|}\\ =0.6275\text{ mile}\end{array}$

Conclusion:

Thus, the required distance of the chain extended is $0.0646\text{ mile}$ , and the chain length is $0.6275\text{ mile}$ .