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Lars Ph
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- Is there a way to use the ewald sphere to calculate h,k,l?
How do I use the ewald sphere to calculate h, k and l?
That will depend on what h, k and l represent.Lars Ph said:How do I use the ewald sphere to calculate h, k and l?
The Ewald sphere is a geometric construct used in crystallography to visualize the relationship between the reciprocal lattice of a crystal and the diffraction pattern produced when X-rays or electrons are scattered by the crystal. It helps in determining the allowed diffraction conditions by representing the reciprocal lattice points in relation to the incident and diffracted beams.
To determine the Miller indices (h, k, l) using the Ewald sphere, you first need to identify the reciprocal lattice points that intersect the Ewald sphere. The coordinates of these points correspond to the Miller indices. You can find these indices by projecting the position of the reciprocal lattice point onto the axes defined by the crystal lattice.
The radius of the Ewald sphere is defined by the wavelength of the incident radiation and is inversely proportional to the momentum transfer during scattering. It determines the range of reciprocal lattice points that can be accessed during diffraction experiments. A larger radius allows for the inclusion of more lattice points, which can lead to a more complete understanding of the crystal structure.
The orientation of the crystal affects the position of the reciprocal lattice points relative to the Ewald sphere. When the crystal is rotated, the reciprocal lattice points also move, which can change which points intersect the Ewald sphere and thus alter the observed diffraction pattern. Proper alignment of the crystal is crucial for accurately determining the Miller indices.
While the Ewald sphere is primarily used in X-ray and electron diffraction, it can also be applied to other scattering methods like neutron diffraction. However, the specific details of the Ewald sphere construction may vary depending on the type of radiation used, as different types of scattering may involve different interactions with the crystal lattice.