Question

copper as FCC structure with the lattice constant 0.362nm calculate the interplanar spacing for 112 plane​

Answer 1

Answer:

copper as FCC structure with the lattice constant 0.362nm calculate the interplanar spacing for 112 plane

Explanation:

copper as FCC structure with the lattice constant 0.362nm calculate the interplanar spacing .

Answer 2

Answer:

Copper as FCC structure with the lattice constant 0.362nm.The interplanar spacing for 112 planes​ will be = $= 1.4778\times 10^{-10} m$

Explanation:

The interplanar spacing or d-spacing is the distance between two adjacent planes of atoms having the same miller indices $\left(d_{h k l}\right)$.

Consider a plane (hkl),

The d- spacing of that cubic material is given by the formula,

$d_{h k l}=\frac{a_{}}{\sqrt{h^{2}+k^{2}+l^{2}}}$  ...(1)

Where a is the lattice parameter.

From the question,

Plane (hkl) = ($112$)

That is, h = 1, k = 1, l = 2

Lattice constant (a) = $0.36$ nm = $0.36\times 10^{-9} \ m$

Substitutes these values into the equation (1),

$d_{h k l}=\frac{0.362\times10^{-9} }{\sqrt{1^{2}+1^{2}+2^{2}}}$

       $=\frac{0.362\times10^{-9} }{\sqrt{6}}$

       $= 1.4778\times 10^{-10} m$

Thus, Interplanar spacing = $= 1.4778\times 10^{-10} m$