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Interplanar Distance in Triclinic 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{((b^{2}) \cdot (c^{2}) \cdot ({(\sin (α))}^{2}) \cdot (h^{2})) + (({a lattice}^{2}) \cdot (c^{2}) \cdot ({(\sin (β))}^{2}) \cdot (k^{2})) + (({a lattice}^{2}) \cdot (b^{2}) \cdot ({(\sin (γ))}^{2}) \cdot (l^{2})) + (2 \cdot a lattice \cdot b \cdot (c^{2}) \cdot ((\cos (α) \cdot \cos (β)) - \cos (γ)) \cdot h \cdot k) + (2 \cdot b \cdot c \cdot ({a lattice}^{2}) \cdot ((\cos (γ) \cdot \cos (β)) - \cos (α)) \cdot l \cdot k) + (2 \cdot a lattice \cdot c \cdot (b^{2}) \cdot ((\cos (α) \cdot \cos (γ)) - \cos (β)) \cdot h \cdot l)}{{V unit cell}^{2}}}}

d - Interplanar Spacing?b - Lattice Constant b?c - Lattice Constant c?α - Lattice parameter alpha?h - Miller Index along x-axis?a_lattice - Lattice Constant a?β - Lattice Parameter Beta?k - Miller Index along y-axis?γ - Lattice Parameter gamma?l - Miller Index along z-axis?V_{unit cell} - Volume of Unit Cell?

Interplanar Distance in Triclinic Crystal Lattice Example

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

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

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

0.0154 = \sqrt{\frac{1}{\frac{((12^{2}) \cdot (15^{2}) \cdot ({(\sin (30))}^{2}) \cdot (9^{2})) + ((14^{2}) \cdot (15^{2}) \cdot ({(\sin (35))}^{2}) \cdot (4^{2})) + ((14^{2}) \cdot (12^{2}) \cdot ({(\sin (38))}^{2}) \cdot (11^{2})) + (2 \cdot 14 \cdot 12 \cdot (15^{2}) \cdot ((\cos (30) \cdot \cos (35)) - \cos (38)) \cdot 9 \cdot 4) + (2 \cdot 12 \cdot 15 \cdot (14^{2}) \cdot ((\cos (38) \cdot \cos (35)) - \cos (30)) \cdot 11 \cdot 4) + (2 \cdot 14 \cdot 15 \cdot (12^{2}) \cdot ((\cos (30) \cdot \cos (38)) - \cos (35)) \cdot 9 \cdot 11)}{105^{2}}}}

0.0154nm = \sqrt{\frac{1}{\frac{(({12A}^{2}) \cdot ({15A}^{2}) \cdot ({(\sin (30°))}^{2}) \cdot (9^{2})) + (({14A}^{2}) \cdot ({15A}^{2}) \cdot ({(\sin (35°))}^{2}) \cdot (4^{2})) + (({14A}^{2}) \cdot ({12A}^{2}) \cdot ({(\sin (38°))}^{2}) \cdot (11^{2})) + (2 \cdot 14A \cdot 12A \cdot ({15A}^{2}) \cdot ((\cos (30°) \cdot \cos (35°)) - \cos (38°)) \cdot 9 \cdot 4) + (2 \cdot 12A \cdot 15A \cdot ({14A}^{2}) \cdot ((\cos (38°) \cdot \cos (35°)) - \cos (30°)) \cdot 11 \cdot 4) + (2 \cdot 14A \cdot 15A \cdot ({12A}^{2}) \cdot ((\cos (30°) \cdot \cos (38°)) - \cos (35°)) \cdot 9 \cdot 11)}{{105A³}^{2}}}}

0.0154 = \sqrt{\frac{1}{\frac{((12^{2}) \cdot (15^{2}) \cdot ({(\sin (30))}^{2}) \cdot (9^{2})) + ((14^{2}) \cdot (15^{2}) \cdot ({(\sin (35))}^{2}) \cdot (4^{2})) + ((14^{2}) \cdot (12^{2}) \cdot ({(\sin (38))}^{2}) \cdot (11^{2})) + (2 \cdot 14 \cdot 12 \cdot (15^{2}) \cdot ((\cos (30) \cdot \cos (35)) - \cos (38)) \cdot 9 \cdot 4) + (2 \cdot 12 \cdot 15 \cdot (14^{2}) \cdot ((\cos (38) \cdot \cos (35)) - \cos (30)) \cdot 11 \cdot 4) + (2 \cdot 14 \cdot 15 \cdot (12^{2}) \cdot ((\cos (30) \cdot \cos (38)) - \cos (35)) \cdot 9 \cdot 11)}{105^{2}}}}

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Interplanar Distance in Triclinic Crystal Lattice Solution

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

FIRST Step Consider the formula

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

Next Step Substitute values of Variables

∴

d = \sqrt{\frac{1}{\frac{(({12A}^{2}) \cdot ({15A}^{2}) \cdot ({(\sin (30°))}^{2}) \cdot (9^{2})) + (({14A}^{2}) \cdot ({15A}^{2}) \cdot ({(\sin (35°))}^{2}) \cdot (4^{2})) + (({14A}^{2}) \cdot ({12A}^{2}) \cdot ({(\sin (38°))}^{2}) \cdot (11^{2})) + (2 \cdot 14A \cdot 12A \cdot ({15A}^{2}) \cdot ((\cos (30°) \cdot \cos (35°)) - \cos (38°)) \cdot 9 \cdot 4) + (2 \cdot 12A \cdot 15A \cdot ({14A}^{2}) \cdot ((\cos (38°) \cdot \cos (35°)) - \cos (30°)) \cdot 11 \cdot 4) + (2 \cdot 14A \cdot 15A \cdot ({12A}^{2}) \cdot ((\cos (30°) \cdot \cos (38°)) - \cos (35°)) \cdot 9 \cdot 11)}{{105A³}^{2}}}}

Next Step Convert Units

∴

d = \sqrt{\frac{1}{\frac{(({1.2E-9m}^{2}) \cdot ({1.5E-9m}^{2}) \cdot ({(\sin (0.5236rad))}^{2}) \cdot (9^{2})) + (({1.4E-9m}^{2}) \cdot ({1.5E-9m}^{2}) \cdot ({(\sin (0.6109rad))}^{2}) \cdot (4^{2})) + (({1.4E-9m}^{2}) \cdot ({1.2E-9m}^{2}) \cdot ({(\sin (0.6632rad))}^{2}) \cdot (11^{2})) + (2 \cdot 1.4E-9m \cdot 1.2E-9m \cdot ({1.5E-9m}^{2}) \cdot ((\cos (0.5236rad) \cdot \cos (0.6109rad)) - \cos (0.6632rad)) \cdot 9 \cdot 4) + (2 \cdot 1.2E-9m \cdot 1.5E-9m \cdot ({1.4E-9m}^{2}) \cdot ((\cos (0.6632rad) \cdot \cos (0.6109rad)) - \cos (0.5236rad)) \cdot 11 \cdot 4) + (2 \cdot 1.4E-9m \cdot 1.5E-9m \cdot ({1.2E-9m}^{2}) \cdot ((\cos (0.5236rad) \cdot \cos (0.6632rad)) - \cos (0.6109rad)) \cdot 9 \cdot 11)}{{1.1E-28m³}^{2}}}}

Next Step Prepare to Evaluate

∴

d = \sqrt{\frac{1}{\frac{(({1.2E-9}^{2}) \cdot ({1.5E-9}^{2}) \cdot ({(\sin (0.5236))}^{2}) \cdot (9^{2})) + (({1.4E-9}^{2}) \cdot ({1.5E-9}^{2}) \cdot ({(\sin (0.6109))}^{2}) \cdot (4^{2})) + (({1.4E-9}^{2}) \cdot ({1.2E-9}^{2}) \cdot ({(\sin (0.6632))}^{2}) \cdot (11^{2})) + (2 \cdot 1.4E-9 \cdot 1.2E-9 \cdot ({1.5E-9}^{2}) \cdot ((\cos (0.5236) \cdot \cos (0.6109)) - \cos (0.6632)) \cdot 9 \cdot 4) + (2 \cdot 1.2E-9 \cdot 1.5E-9 \cdot ({1.4E-9}^{2}) \cdot ((\cos (0.6632) \cdot \cos (0.6109)) - \cos (0.5236)) \cdot 11 \cdot 4) + (2 \cdot 1.4E-9 \cdot 1.5E-9 \cdot ({1.2E-9}^{2}) \cdot ((\cos (0.5236) \cdot \cos (0.6632)) - \cos (0.6109)) \cdot 9 \cdot 11)}{{1.1E-28}^{2}}}}

Next Step Evaluate

∴

d = 1.53891539382534E-11m

Next Step Convert to Output's Unit

∴

d = 0.0153891539382534nm

LAST Step Rounding Answer

∴

d = 0.0154nm

Interplanar Distance in Triclinic 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.

Formulas to find Interplanar Spacing

Lattice Constant b

The Lattice Constant b refers to the physical dimension of unit cells in a crystal lattice along y-axis.

Symbol: b

Measurement: LengthUnit: A