Calculating Interplanar Spacing in Orthorhombic Unit Cell

The question asks to determine the separation (interplanar spacing) for the {123} and {246} planes within an orthorhombic unit cell. We are given the lattice parameters of the cell.

The given information is:

• Lattice parameter a = 0.82 nm
• Lattice parameter b = 0.94 nm
• Lattice parameter c = 0.75 nm
• Plane 1: {123} (Miller indices h=1, k=2, l=3)
• Plane 2: {246} (Miller indices h=2, k=4, l=6)

Orthorhombic Interplanar Spacing Formula

For an orthorhombic crystal system, the interplanar spacing \( d_{hkl} \) between planes with Miller indices (hkl) is given by the formula:

\( \frac{1}{d_{hkl}^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2} \)

This can be rearranged to solve for \( d_{hkl} \):

\( d_{hkl} = \frac{1}{\sqrt{\frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2}}} \)

Separation of {123} Plane Calculation

For the {123} plane, h=1, k=2, and l=3. We use the given lattice parameters a=0.82 nm, b=0.94 nm, and c=0.75 nm in the formula:

\( d_{123} = \frac{1}{\sqrt{\frac{1^2}{(0.82)^2} + \frac{2^2}{(0.94)^2} + \frac{3^2}{(0.75)^2}}} \)

\( d_{123} = \frac{1}{\sqrt{\frac{1}{0.6724} + \frac{4}{0.8836} + \frac{9}{0.5625}}} \)

\( d_{123} = \frac{1}{\sqrt{1.4872 + 4.5269 + 16.0000}} \)

\( d_{123} = \frac{1}{\sqrt{22.0141}} \)

\( d_{123} \approx \frac{1}{4.692} \)

\( d_{123} \approx 0.2131 \text{ nm} \)

Rounding this value, the separation for the {123} plane is approximately 0.21 nm.

Separation of {246} Plane Calculation

For the {246} plane, h=2, k=4, and l=6. We use the same lattice parameters a=0.82 nm, b=0.94 nm, and c=0.75 nm:

\( d_{246} = \frac{1}{\sqrt{\frac{2^2}{(0.82)^2} + \frac{4^2}{(0.94)^2} + \frac{6^2}{(0.75)^2}}} \)

\( d_{246} = \frac{1}{\sqrt{\frac{4}{0.6724} + \frac{16}{0.8836} + \frac{36}{0.5625}}} \)

\( d_{246} = \frac{1}{\sqrt{5.9497 + 18.1078 + 64.0000}} \)

\( d_{246} = \frac{1}{\sqrt{88.0575}} \)

\( d_{246} \approx \frac{1}{9.384} \)

\( d_{246} \approx 0.1066 \text{ nm} \)

Rounding this value, the separation for the {246} plane is approximately 0.11 nm.

Alternatively, we can observe that the Miller indices (246) are proportional to (123), where (246) = 2*(123). For proportional indices (nh, nk, nl), the interplanar spacing \( d_{nh,nk,nl} \) is related to \( d_{hkl} \) by \( d_{nh,nk,nl} = \frac{d_{hkl}}{n} \). In this case, n=2.

So, \( d_{246} = \frac{d_{123}}{2} \approx \frac{0.2131}{2} \approx 0.10655 \text{ nm} \), which also rounds to 0.11 nm.

Final Separation Values

The separation of the {123} plane is approximately 0.21 nm, and the separation of the {246} plane is approximately 0.11 nm.

Thus, the separations are 0.21 nm and 0.11 nm respectively.