Discussion
In a simple cubic crystal of lattice spacing a = 3 Å, a positive edge dislocation 1 mm long climbs down by 1 \(\mu\)m. The number of vacancies created or lost in the crystal is :
Answer: A
The number of vacancies is 1.10 × 10\(^{10}\).
When an edge dislocation climbs, the edge of its extra half-plane gains or loses a row of atoms for every lattice spacing it moves. Climbing down extends the half-plane, and each added atom leaves a vacancy behind. The number of atoms per row and the number of rows are:
\[ \begin{aligned} \frac{L}{a} &= \frac{10^{-3}}{3 \times 10^{-10}} = 3.33 \times 10^{6} \\ \frac{h}{a} &= \frac{10^{-6}}{3 \times 10^{-10}} = 3.33 \times 10^{3} \end{aligned} \]
Their product is \(1.11 \times 10^{10}\), option A. The larger options would need the climb to be counted over an area or volume much greater than the plane swept by the dislocation line.
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