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 \(10^{-9}\ \text{J/m}\).
Show Hint
Find the FCC Burgers vector \(b=a/\sqrt{2}\), then use \(E/L=\tfrac{1}{2}Gb^2\).
Step 1: Write the strain energy relation to use.
For a screw dislocation, the elastic energy stored per unit length is taken as
\[ \frac{E}{L}=\frac{1}{2}Gb^2 \]
so we need $G$ and the Burgers vector $b$ before anything else.
Step 2: Get the Burgers vector from the FCC slip vector.
FCC crystals slip along $\tfrac{a}{2}\langle110\rangle$ directions. A $\langle110\rangle$ direction like $[110]$ has two unit components, so its length is $\sqrt{1^2+1^2}=\sqrt2$, giving
\[ b=\frac{a}{2}\times\sqrt2=\frac{a}{\sqrt2} \]
With $a=0.35$ nm,
\[ b=\frac{0.35\times10^{-9}}{\sqrt2}=2.475\times10^{-10}\ \text{m} \]
Step 3: Build the product $Gb^2$ as one block.
\[ Gb^2=(76\times10^{9})\times(2.475\times10^{-10})^2 \]
\[ =(76\times10^{9})\times(6.125\times10^{-20}) \]
\[ =4.655\times10^{-9}\ \text{J/m} \]
Step 4: Halve it to get the final energy per length.
\[ \frac{E}{L}=\frac{Gb^2}{2}=\frac{4.655\times10^{-9}}{2}=2.3275\times10^{-9}\ \text{J/m} \]
Step 5: Round to two decimals.
$2.3275$ rounds to $2.33$, so the strain energy per unit length is $2.33\times10^{-9}$ J/m, which sits inside the accepted band of $2.30$ to $2.38$.
\[ \boxed{\dfrac{E}{L}\approx2.33\times10^{-9}\ \text{J/m}} \]
