Question

A single crystal BCC metal with a lattice parameter $a = 0.4$ nm is subjected to deformation at a shear strain rate of 0.001 s$^{-1$. If the average mobile dislocation density in the single crystal is $10^{10}$ m$^{-2}$, the average dislocation velocity is ............. × 10$^{-3}$ m s$^{-1}$ (rounded off to two decimal places). Given: Burgers vector $b = \dfrac{a}{2}\langle 111 \rangle$.}

Hint:

Always apply Orowan's relation $\dot{\gamma} = \rho b v$ for dislocation motion. Carefully compute Burgers vector for the given crystal structure.

Answer 1

Step 1: Write Orowan's equation.
The relation between shear strain rate, dislocation density, Burgers vector and dislocation velocity is: \[ \dot{\gamma} = \rho \, b \, v \] where $\dot{\gamma}$ = shear strain rate, $\rho$ = mobile dislocation density, $b$ = Burgers vector magnitude, $v$ = dislocation velocity. Step 2: Calculate Burgers vector.
For BCC: \[ b = \frac{a}{2}\sqrt{1^2 + 1^2 + 1^2} = \frac{a}{2}\sqrt{3} \] Given $a = 0.4$ nm = $0.4 \times 10^{-9}$ m. \[ b = \frac{0.4 \times 10^{-9}}{2} \sqrt{3} = 0.2 \times 10^{-9} \times 1.732 \] \[ b = 0.346 \times 10^{-9} \, m \] Step 3: Rearrange Orowan's equation.
\[ v = \frac{\dot{\gamma}}{\rho b} \] Step 4: Substitute values.
\[ v = \frac{0.001}{10^{10} \times 0.346 \times 10^{-9}} \] \[ v = \frac{0.001}{3.46} \] \[ v = 2.89 \times 10^{-4} \, m s^{-1} \] Now, express as ×10$^{-3}$: \[ v = 0.289 \times 10^{-3} \, m s^{-1} \] But wait – let's recheck carefully: \[ 10^{10} \times 0.346 \times 10^{-9} = 3.46 \] \[ 0.001 / 3.46 = 2.89 \times 10^{-4} \] So final value: \[ \boxed{0.29 \times 10^{-3} \, m s^{-1}} \]