Reciprocal Lattices to SC, FCC and BCC How would you find the Reciprocal Lattices to SC, FCC and BCC Primitive Direct lattice Reciprocal lattice Volume of RL SC BCC FCC Direct Reciprocal Simple cubic bcc fcc Got here (slide 19) in talkative class. Spend a lot of time talking about Perovskites to review basic crystal structure concepts. If we took b1 dot (b2 cross b3) we’d get the volume of the reciprocal cell, which would give these. Might come back and prove these values if time at the end of class. Makes sense since real space volume was smaller for FCC in real space, so bigger in reciprocal space.

Volume of the Brillouin Zone (BZ) In general the volume of the BZ is equal to (2 )3 Volume of real space primitive lattice Specifically go through BCC (easiest after sc). Mention if take b1 dot (b2 x b3) is how you get the last column on the reciprocal lattice slide. While their volumes are the same, they are not exactly the same thing. Remember that there were many ways to define a primitive unit cell in real space, and the Wigner Seitz cell was a specific way to generate a unit cell (by drawing perpendicular bisectors to neighboring lattice points). The same thing is occurring here, as we’ll learn on the next slide, since the BZ is the Wigner Seitz cell of the reciprocal lattice. Thus the volume of the reciprocal lattice is equal to the volume of the BZ.

Discuss the reciprocal lattice in 1D Real lattice x What is the range of unique environments? -/a /a Reciprocal lattice Look familiar? k -6/a -4/a -2/a 2/a 4/a Wigner Seitz Cell: Smallest space enclosed when intersecting the midpoint to the neighboring lattice points. Why don’t we include second neighbors here (do in 2D/3D)?

The Brillouin Zone -/a /a Reciprocal lattice k 4/a -6/a -4/a -2/a 2/a Is defined as the Wigner-Seitz primitive cell in the reciprocal lattice (smallest volume/area/distance in RL) Its construction exhibits all the wavevectors k which can be Bragg-reflected by the crystal Got here in 50 minute class, XRD pattern took some time

Group: Draw the 1st Brillouin Zone of a sheet of graphene Only overlaying the grid as a visual aid (not part of the lattice). a1* a1 a2 a2* Note that this definition of the real space unit cell is a little different from the other one I’ve used. There actually aren’t any atoms at the lattice point. Message: There are lots of different ways to define a unit cell. Regardless, the BZ will look the same. Real Space 2-atom basis In what directions do b1 and b2 point? Wigner-Seitz Unit Cell of Reciprocal Lattice = First Brillouin zone

Group: Polonium Consider simple cubic polonium, Po, which is similar to a 1D chain in 3 dimensions. (a) Determine the lengths and directions of the lattice translation vectors for the lattice which is reciprocal to the real-space Po lattice. (b) The first Brillouin Zone is defined to be the Wigner-Seitz primitive cell of the reciprocal lattice. Sketch the first Brillouin Zone of Po. (Identify the location of the faces of the shape.) (c) Show that the volume of the first Brillouin Zone is (2)3/V , where V is the volume of the real space primitive unit cell. For the group project, I’m most interested in you doing the red part, but the rest might be useful to help you get there and/or think about the result. 6N1p Full problem (I have shortened this problem to get us to the point faster): (a) Taking a Po atom as a lattice point, construct the Wigner-Seitz cell of polonium in real space. What is it’s volume? (b) Work out the lengths and directions of the lattice translation vectors for the lattice which is reciprocal to the real-space Po lattice. (c) The first Brillouin Zone is defined to be the Wigner-Seitz primitive cell of the reciprocal lattice. Sketch the first Brillouin Zone of Po. (d) Show that the volume of the first Brillouin Zone is (2)3/V , where V is the volume of the real space primitive unit cell. (If I use a problem like this in writing a book, strip off the atom type (not many people care about Po). Double check that both O and F arrange in this structure and maybe focus on that instead.)

Square Lattice (on board) Introduction of Higher Order BZs (HOBZs will seem more important in Ch.9)

Group: Determine the shape of the BZ of the FCC Lattice How many sides will it have and along what directions? How might you approach this? SC BCC FCC # of nearest neighbors 6 8 12 Nearest-neighbor distance a ½ a 3 a/2 # of second neighbors Second neighbor distance a2 FCC Primitive and Conventional Unit Cells

Reciprocal Space to the FCC Lattice Bring print out of the shapes for them to think about

WS zone and BZ The BZ of fcc is the WS cell of bcc. Lattice Real Space Lattice K-space bcc WS cell Bcc BZ (fcc lattice in K-space) fcc WS cell fcc BZ (bcc lattice in K-space) The WS cell of bcc lattice in real space transforms to a Brillouin zone in a fcc lattice in reciprocal space while the WS cell of a fcc lattice transforms to a Brillouin zone of a bcc lattice in reciprocal space. (But this is a lot of words to put up on a slide, so I shortened it to what is on the bottom of the slide.) The BZ of fcc is the WS cell of bcc. The BZ of bcc is the WS cell of fcc.

Labelling the BZ Usually, it is sufficient to know the energy En(k) curves - the dispersion relations - along the major directions. Directions are chosen that lead along special symmetry points. These points are labeled according to the following rules: Direction along BZ We are moving in the direction of being able to calculate these energy bands, but first it can be useful just to know what they refer to. Points on the surface (red) of the Brillouin zone are Roman letters. Points (and lines) inside the Brillouin zone are denoted with Greek letters. The center of the Wigner-Seitz cell is always denoted by a G Energy or Frequency

Brillouin Zones in 3D fcc bcc hcp The BZ reflects lattice symmetry Note: bcc lattice in reciprocal space is a fcc lattice hcp Note: fcc lattice in reciprocal space is a bcc lattice The BZ reflects lattice symmetry Construction leads to primitive unit cell in rec. space

Brillouin Zone of Silicon Symbol Description Γ Center of the Brillouin zone Simple Cubic M Center of an edge R Corner point X Center of a face FCC K Middle of an edge joining two hexagonal faces L Center of a hexagonal face C6 U Middle of an edge joining a hexagonal and a square face W Center of a square face C4 BCC H Corner point joining 4 edges N P Corner point joining 3 edges What kind of crystal structure is Si? Why do I say semiconductors? Metals are most freely so barely effected. Insulators don’t really conduct so don’t move around. Points of symmetry on the BZ are important (e.g. determining bandstructure). Electrons in semiconductors are perturbed by the potential of the crystal, which varies across unit cell.

Learning Objectives for Diffraction After our diffraction class you should be able to: Explain why diffraction occurs Utilize Bragg’s law to determine angles of diffraction Briefly discuss some different diffraction techniques (Next time) Determine the lattice type and lattice parameters of a material given an XRD pattern and the x-ray energy Alternative reference: Ch. 2 Kittel

Continuum limit: Where the wavelength is bigger than the spacing between atoms. Otherwise diffraction effects dominate. Why might you want to think about optical properties? There are at least two reasons. One is that you can make use of known optical materials to design and build devices to manipulate light: mirrors, lenses, filters, polarizers, and a host of others. The second is that you can measure the optical response of a new material and obtain a wealth of information about the low energy excitations in the solid. With the use of your own eyes, you could see that solids have a wide range of optical properties. Silver is a lustrous metal, with a high reflectance over the whole visible range. Silicon is a crystalline semiconductor and the basis of modern electronics. With the surface oxide freshly etched off, silicon is also rather reflective, although not as good a mirror as silver. ∗ Salt (sodium chloride) is a transparent ionic insulator, is necessary for life, and makes up about 3.5% (by weight) of seawater. A crystal of salt is transparent over the entire visible spectrum; because the refractive index is about 1.5, the reflectance is about 4%. If you had ultraviolet eyes, you would see these materials differently. Silver would be a poor reflector, with at most 20% reflectance and trailing off to zero at the shortest wavelengths. In contrast, the reflectance of silicon would be better than in the visible, reaching up to 75%. Sodium chloride would be opaque over much of the spectrum, with a reflectance a bit higher than in the visible. Those with infrared eyes would also see things differently from visible or uv-sensitive individuals. Silver would have a reflectance above 99%. Silicon would appear opaque at the shortest infrared wavelengths but would then become transparent, so that you could see through even meter-thick crystals.† Sodium chloride remains transparent over much of the infrared

Application of XRD XRD is a nondestructive and cheap technique. Some of the uses of x-ray diffraction are: Determination of the structure of crystalline materials Determination of the orientation of single crystals Differentiation between crystalline and amorphous materials Determination of the texture of polygrained materials Measurement of layer thickness Measurement of (epitaxial) strain Determination of electron distribution within the atoms, and throughout the unit cell Need to get rid of old duplicates following this.

DIFFRACTION Diffraction is a wave phenomenon in which the apparent bending and spreading of waves when they meet an obstruction is measured. Diffraction occurs with electromagnetic waves, such as light and radio waves, and also in sound waves and water waves. X-ray diffraction is optimally sensitive to the periodic nature of the solid’s atomic structure.

When X-rays interact with atoms, you get scattering Scattering is the emission of X-rays of the same frequency/energy as the incident X-rays in all directions (but with much lower intensity) Similar to the double slit experiment, this scattering will sometimes be constructive Second order process due to phase ambiguity, doesn’t matter if initial photon is absorbed before emitted photon. See from Fermi’s golden rule (Ch.4 in the book).

Will look at this again shortly Incident beam Will look at this again shortly Second order Note diameter of smallest circle is one wavelength. Third circular is two wavelengths. The light must be incident over some time to create the wavefronts shown. Zeroth Order

Physical Model for X-ray Scattering Consider a plane wave scattering on an atom. Atom What isn’t r squared? We are looking at psi, not intensity. This is a different approach from the book, but I think its helpful for visualizing the geometry involved.

Diffraction In a Crystal Generic incoming radiation amplitude is: To calculate amplitude of scattered waves at detector position, sum over contributions of all scattering centers Pi with scattering amplitude (form factor) f: Pi ri ko R R’-ri R’ source R, R’ >> ri Generic for any point in the sample. (ignoring the time dependence, will disappear when we multiply by the complex conjugate) If R is much greater than ri, then can pull out the exponents involving R from both si incident and si outgoing Detector The intensity that is measured (can’t measure amplitude) is Scattering vector The book calls K, but G is another common notation.

Diffraction Theory k’ K=k’-ko ko Pi ri ko ko R R’-ri R’ source Generic for any point in the sample. (ignoring the time dependence, will disappear when we square it) If R is much greater than ri, then can pull out the exponents involving R from both si incident and si outgoing Detector The intensity that is measured (can’t measure amplitude) is Scattering vector The book calls K, but G is another common notation.

How does this limit ? where, d is the spacing of the planes and n is the order of diffraction. Bragg reflection can only occur for wavelength This is why we cannot use visible light. No diffraction occurs when the above condition is not satisfied. What jumps out at you as a necessary condition for this to be true?

Non-xray Diffraction Methods (more in later chapters) Any particle will scatter and create diffraction pattern Beams are selected by experimentalists depending on sensitivity X-rays not sensitive to low Z elements, but neutrons are Electrons sensitive to surface structure if energy is low Atoms (e.g., helium) sensitive to surface only For inelastic scattering, momentum conservation is important X-Ray Neutron Electron Replace top figure (not so informative, probably more confusing) λ = 1A° E ~ 104 eV interact with electron Penetrating λ = 1A° E ~ 0.08 eV interact with nuclei Highly Penetrating λ = 2A° E ~ 150 eV interact with electron Less Penetrating