Question

The lattice parameter of Ni (face centered cubic) is \(0.35\) nm and its shear modulus is \(76\) GPa. Find the strain energy per unit length of a screw dislocation in the Ni crystal (rounded off to two decimal places), in units of \(10^{-9}\ \text{J/m}\).

Hint:

Find the FCC Burgers vector \(b=a/\sqrt{2}\), then use \(E/L=\tfrac{1}{2}Gb^2\).

Answer 1

Correct Answer: 2.3

Step 1: Recall the Burgers vector for an FCC lattice.
In a face centered cubic (FCC) crystal, dislocations glide along the closest packed direction, which belongs to the \(\langle 110 \rangle\) family. The Burgers vector for this slip system has magnitude
\[ b = \frac{a}{\sqrt{2}} \]
where \(a\) is the lattice parameter.

Step 2: Compute the Burgers vector for Ni.
Given \(a = 0.35\) nm,
\[ b = \frac{0.35}{\sqrt{2}} = 0.2475\ \text{nm} = 2.475 \times 10^{-10}\ \text{m} \]

Step 3: Recall the strain energy formula for a screw dislocation.
The elastic strain energy stored per unit length of a dislocation is commonly estimated as
\[ \frac{E}{L} = \frac{1}{2}Gb^2 \]
where \(G\) is the shear modulus and \(b\) is the Burgers vector magnitude. This is the standard order of magnitude form used for dislocation strain energy, arrived at after folding the outer to inner cut off radius ratio into the constant.

Step 4: Square the Burgers vector.
\[ b^2 = (2.475\times10^{-10})^2 = 6.125\times10^{-20}\ \text{m}^2 \]

Step 5: Substitute the shear modulus and compute.
Given \(G = 76\) GPa \(= 76\times10^{9}\) Pa,
\[ \frac{E}{L} = \frac{1}{2}\times76\times10^{9}\times6.125\times10^{-20} \]
\[ = 0.5 \times 4.655\times10^{-9} \]
\[ = 2.3275\times10^{-9}\ \text{J/m} \]

Final Answer:
Rounded to two decimal places, the strain energy per unit length of the screw dislocation is 2.33 (in units of \(10^{-9}\) J/m).
\[ \boxed{\frac{E}{L} \approx 2.33\times10^{-9}\ \text{J/m}} \]