Question

The separation (in nm) of {134} planes of an orthorhombic unit cell (with cell parameters 𝑎=0.5 nm, 𝑏=0.6 nm, and 𝑐=0.8 nm) is

• A. 0.036
• B. 0.136
• C. 0.236
• D. 0.336

Answer 1

The Correct Option is B

The task is to determine the separation, also known as the interplanar spacing, of the {134} planes in an orthorhombic unit cell with given cell parameters. To solve this, we will use the formula for interplanar spacing for an orthorhombic lattice:

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

Where:

• \(h, k, l\) are the Miller indices of the plane. 
• \(a, b, c\) are the lattice parameters.

Given:

• h = 1, k = 3, l = 4
• a = 0.5 nm, b = 0.6 nm, c = 0.8 nm

Substitute these values into the formula:

Calculate each term in the denominator:

• \(\frac{1^2}{a^2} = \frac{1}{(0.5)^2} = \frac{1}{0.25} = 4\)
• \(\frac{3^2}{b^2} = \frac{9}{(0.6)^2} = \frac{9}{0.36} = 25\)
• \(\frac{4^2}{c^2} = \frac{16}{(0.8)^2} = \frac{16}{0.64} = 25\)

Add these values:

\(4 + 25 + 25 = 54\)

Now, calculate the interplanar spacing:

\(d_{134} = \frac{1}{\sqrt{54}}\)

Compute \(\sqrt{54}\):

\(\sqrt{54} \approx 7.348\)

Thus:

\(d_{134} = \frac{1}{7.348} \approx 0.136 \text{ nm}\)

Therefore, the separation of the {134} planes is approximately 0.136 nm.

The correct answer is: 0.136