The question asks for the magnitude of the Burgers vector in copper, given its lattice parameter. Copper has a Face-Centered Cubic (FCC) crystal structure.
Determining Burgers Vector Magnitude for FCC Copper
For FCC crystals, the shortest Burgers vector (b) lies along the $ \langle 110 \rangle $ direction and its magnitude is half the length of the face diagonal.
- Lattice parameter of copper, $ a = 3.61 \text{ Å} $.
- The face diagonal length in an FCC lattice is $ d = a\sqrt{2} $.
- The Burgers vector magnitude is $ b = \frac{d}{2} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}} $.
Calculation
Substitute the given lattice parameter into the formula:
$ b = \frac{3.61 \text{ Å}}{\sqrt{2}} $ $ b \approx \frac{3.61 \text{ Å}}{1.4142} $ $ b \approx 2.5526 \text{ Å} $Conclusion
The calculated magnitude is approximately $ 2.55 \text{ Å} $. Comparing this with the given options, the closest value is $ 2.54 \text{ Å} $.
The shear modulus value provided is not needed for this calculation.
Important Questions from Defects Dislocation Stress Field Burgers Vector
- Which one of the following dislocation dissociation reactions is feasible in face-centered cubic metals?
- With reference to edge and screw dislocations, which of the following statements is/are CORRECT?
The Burger's vector of a dislocation in a cubic crystal (with lattice parameter a) is $\frac{a}{2}[110]$ and dislocation line is along $[112]$ direction. The angle (in degrees) between the dislocation line and its Burger's vector is _________
- A plastically deformed metal crystal at low temperature exhibits wavy slip line pattern due to
- The c/a ratio of Zn (hcp) is 1.856. Slip at room temperature occurs most easily on which of the following slip systems in Zn:
Note: In hcp metals, the ideal c/a ratio is 1.633.
