Orthorhombic Crystal Interplanar Separation Calculation
This solution explains how to calculate the interplanar separation for the (110) planes in an orthorhombic crystal with specific lattice parameter relationships.
Understanding Orthorhombic System Properties
- In an orthorhombic crystal system, the lattice axes are mutually perpendicular ($\alpha = \beta = \gamma = 90^\circ$) but have unequal lengths ($a \neq b \neq c$).
- The lattice parameter relations are given as $a = 2b = 3c$.
- From these relations, we can express $b$ and $c$ in terms of $a$:
- $b = \frac{a}{2}$
- $c = \frac{a}{3}$
Interplanar Spacing Formula for Orthorhombic System
The formula for the interplanar spacing ($d_{hkl}$) between planes (hkl) in an orthorhombic crystal is:
$ \frac{1}{d_{hkl}^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2} $Calculating Separation for (110) Planes
- Substitute Miller Indices: For the (110) planes, the Miller indices are $h=1$, $k=1$, and $l=0$.
- Apply Formula: Substitute these values into the interplanar spacing formula: $ \frac{1}{d_{110}^2} = \frac{1^2}{a^2} + \frac{1^2}{b^2} + \frac{0^2}{c^2} $ $ \frac{1}{d_{110}^2} = \frac{1}{a^2} + \frac{1}{b^2} $
- Express Lattice Parameters in Units of a: Substitute $b = \frac{a}{2}$: $ \frac{1}{d_{110}^2} = \frac{1}{a^2} + \frac{1}{(\frac{a}{2})^2} $ $ \frac{1}{d_{110}^2} = \frac{1}{a^2} + \frac{1}{\frac{a^2}{4}} $ $ \frac{1}{d_{110}^2} = \frac{1}{a^2} + \frac{4}{a^2} $ $ \frac{1}{d_{110}^2} = \frac{5}{a^2} $
- Solve for $d_{110}$: Rearrange the equation to find $d_{110}$: $ d_{110}^2 = \frac{a^2}{5} $ $ d_{110} = \sqrt{\frac{a^2}{5}} = \frac{a}{\sqrt{5}} $
- Calculate Value in Units of a: To express the interplanar separation in units of $a$, calculate $\frac{d_{110}}{a}$: $ \frac{d_{110}}{a} = \frac{1}{\sqrt{5}} $ Using a calculator, $\sqrt{5} \approx 2.2360679$. $ \frac{d_{110}}{a} \approx \frac{1}{2.2360679} \approx 0.44721359... $
- Round to Three Decimal Places: Rounding the result to three decimal places gives $0.447$.
The calculated value $\frac{d_{110}}{a} \approx 0.447$ lies within the specified range of 0.445 to 0.45.
Important Questions from Crystal Structure Bravais Lattices Unit Cell
- For a two-dimensional hexagonal lattice with lattice constant $ a $, the atomic density is
- Consider a crystal that has a basis of one atom. Its primitive vectors are $ \vec{a_1} = a\hat{i} $, $ \vec{a_2} = a\hat{j} $, $ \vec{a_3} = \frac{a}{2}(\hat{i} + \hat{j} + \hat{k}) $, where $ \hat{i}, \hat{j}, \hat{k} $ are the unit vectors in the $ x, y $ and $ z $ directions of the Cartesian coordinate system and $ a $ is a positive constant. Which one of the following is the correct option regarding the type of the Bravais lattice?
- A compound consists of three ions X, Y and Z. The Z ions are arranged in an FCC arrangement. The X ions occupy $\frac{1}{6}$ of the tetrahedral voids and the Y ions occupy $\frac{1}{3}$ of the octahedral voids. Which one of the following is the CORRECT chemical formula of the compound?
For the given unit cells of a two dimensional square lattice, which option lists all the primitive cells?
The number of distinct ways the primitive unit cell can be constructed for the two dimensional lattice as shown in the figure is ______.
