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Interplanar Distance in Monoclinic Crystal Lattice Formula

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Interplanar Spacing is the distance between adjacent and parallel planes of the crystal. Check FAQs

d = \sqrt{\frac{1}{\frac{(\frac{h^{2}}{{a lattice}^{2}}) + (\frac{(k^{2}) \cdot ({\sin (β)}^{2})}{b^{2}}) + (\frac{l^{2}}{c^{2}}) - (2 \cdot h \cdot l \cdot \frac{\cos (β)}{a lattice \cdot c})}{{(\sin (β))}^{2}}}}

d - Interplanar Spacing?h - Miller Index along x-axis?a_lattice - Lattice Constant a?k - Miller Index along y-axis?β - Lattice Parameter Beta?b - Lattice Constant b?l - Miller Index along z-axis?c - Lattice Constant c?

Interplanar Distance in Monoclinic Crystal Lattice Example

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Here is how the Interplanar Distance in Monoclinic Crystal Lattice equation looks like with Values.

Here is how the Interplanar Distance in Monoclinic Crystal Lattice equation looks like with Units.

Here is how the Interplanar Distance in Monoclinic Crystal Lattice equation looks like.

0.1236 = \sqrt{\frac{1}{\frac{(\frac{9^{2}}{14^{2}}) + (\frac{(4^{2}) \cdot ({\sin (35)}^{2})}{12^{2}}) + (\frac{11^{2}}{15^{2}}) - (2 \cdot 9 \cdot 11 \cdot \frac{\cos (35)}{14 \cdot 15})}{{(\sin (35))}^{2}}}}

0.1236nm = \sqrt{\frac{1}{\frac{(\frac{9^{2}}{{14A}^{2}}) + (\frac{(4^{2}) \cdot ({\sin (35°)}^{2})}{{12A}^{2}}) + (\frac{11^{2}}{{15A}^{2}}) - (2 \cdot 9 \cdot 11 \cdot \frac{\cos (35°)}{14A \cdot 15A})}{{(\sin (35°))}^{2}}}}

0.1236 = \sqrt{\frac{1}{\frac{(\frac{9^{2}}{14^{2}}) + (\frac{(4^{2}) \cdot ({\sin (35)}^{2})}{12^{2}}) + (\frac{11^{2}}{15^{2}}) - (2 \cdot 9 \cdot 11 \cdot \frac{\cos (35)}{14 \cdot 15})}{{(\sin (35))}^{2}}}}

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Home » Chemistry » Solid State Chemistry » Inter Planar Distance and Inter Planar Angle » Interplanar Distance in Monoclinic Crystal Lattice

Interplanar Distance in Monoclinic Crystal Lattice Solution

Follow our step by step solution on how to calculate Interplanar Distance in Monoclinic Crystal Lattice?

FIRST Step Consider the formula

d = \sqrt{\frac{1}{\frac{(\frac{h^{2}}{{a lattice}^{2}}) + (\frac{(k^{2}) \cdot ({\sin (β)}^{2})}{b^{2}}) + (\frac{l^{2}}{c^{2}}) - (2 \cdot h \cdot l \cdot \frac{\cos (β)}{a lattice \cdot c})}{{(\sin (β))}^{2}}}}

Next Step Substitute values of Variables

∴

d = \sqrt{\frac{1}{\frac{(\frac{9^{2}}{{14A}^{2}}) + (\frac{(4^{2}) \cdot ({\sin (35°)}^{2})}{{12A}^{2}}) + (\frac{11^{2}}{{15A}^{2}}) - (2 \cdot 9 \cdot 11 \cdot \frac{\cos (35°)}{14A \cdot 15A})}{{(\sin (35°))}^{2}}}}

Next Step Convert Units

∴

d = \sqrt{\frac{1}{\frac{(\frac{9^{2}}{{1.4E-9m}^{2}}) + (\frac{(4^{2}) \cdot ({\sin (0.6109rad)}^{2})}{{1.2E-9m}^{2}}) + (\frac{11^{2}}{{1.5E-9m}^{2}}) - (2 \cdot 9 \cdot 11 \cdot \frac{\cos (0.6109rad)}{1.4E-9m \cdot 1.5E-9m})}{{(\sin (0.6109rad))}^{2}}}}

Next Step Prepare to Evaluate

∴

d = \sqrt{\frac{1}{\frac{(\frac{9^{2}}{{1.4E-9}^{2}}) + (\frac{(4^{2}) \cdot ({\sin (0.6109)}^{2})}{{1.2E-9}^{2}}) + (\frac{11^{2}}{{1.5E-9}^{2}}) - (2 \cdot 9 \cdot 11 \cdot \frac{\cos (0.6109)}{1.4E-9 \cdot 1.5E-9})}{{(\sin (0.6109))}^{2}}}}

Next Step Evaluate

∴

d = 1.23627623337386E-10m

Next Step Convert to Output's Unit

∴

d = 0.123627623337386nm

LAST Step Rounding Answer

∴

d = 0.1236nm

Interplanar Distance in Monoclinic Crystal Lattice Formula Elements

Variables

Functions

Interplanar Spacing

Interplanar Spacing is the distance between adjacent and parallel planes of the crystal.

Symbol: d

Measurement: WavelengthUnit: nm

Note: Value should be greater than 0.

Miller Index along x-axis

The Miller Index along x-axis form a notation system in crystallography for planes in crystal (Bravais) lattices along the x-direction.

Symbol: h

Measurement: NAUnit: Unitless