Bragg Diffraction from 2-D Nanoparticle Arrays adapted by Preston Snee and Ali Jawaid

Transcription

1 ADH 9/9/2013 I. Introduction Bragg Diffraction from 2-D Nanoparticle Arrays adapted by Preston Snee and Ali Jawaid In this module, we will investigate how a crystal lattice of nanoparticles can act as a diffraction grating. The nanoparticles can scatter coherent light from a laser to create a diffraction pattern from which their 2-dimensional structure can be analyzed. Because the microspheres pack into a hexagonal lattice, the distance between each microsphere (and thus the microsphere diameter) will be determined using standard trigonometric relationships. In the 17 th and 18 th centuries, the nature of how light interacts with a periodic array of atoms (ie. a crystal) was becoming of significant interest; in particular, the idea that an atomic crystal lattice can act as a diffraction grating for a small wavelengths of light (such as X-rays) to resolve the structure of salts, metals, and semiconductors. During the latter half of the 20 th century, the role of crystallography was cemented after Rosalyn Franklin s pioneering work largely contributed to elucidating the 3-dimensional structure of the DNA double helix. This phenomenon, called Bragg diffraction, occurs when an incident light source interacts with a periodic or crystalline substrate to form an interference pattern from elastically scattered light waves. The pattern can be analyzed to measure atomic distances (Fig. 1). Traditionally, the diffraction of X-rays has been used to investigate the atomic structure of proteins, RNA, DNA, and inorganic materials as atomic bond lengths are on the order of a typical X-ray (~150 pm). In general, the incident wavelength must be on the order of the distances being measured. In a periodic array with a distance d between successive planes, the Bragg equation describes the locations in which the radiation interferes constructively with itself: 2dsinθ = nλ (1) The wavelength of incident radiation is λ, the deflection angle is θ, and n is an integer. To elaborate on this concept, a representation of light interacting with a periodic lattice is shown in Fig. 1 (left) below. In this scheme, the 1 st ray incident on atom x has the same wavelength as the 2 nd ray incident on atom y; they are also perfectly parallel up to the point where the 1 st ray strikes atom x and is scattered off it. The 2 nd beam continues onwards to the next layer where it is scattered by atom y. The two rays will interfere constructively if the difference in path lengths, after scattering out of the crystal, is an integer multiple of their wavelength.

2 Figure 1. Two light rays are incident on a crystal surface. The conditions for Bragg diffraction requires the two rays to be in phase to produce an interference pattern that will allow one to determine the distance (d) between the two atoms, x and y. As shown in Fig. 1 (right), perpendicular lines off of the 2 nd ray to point x (dashed lines) from points a & b indicate the path difference traveled by the 2 nd ray, which are the line segments ay and yb. This distance: ay + yb = 2 ay (2) needs to be an integer multiple of the wavelength (λ) if the rays are to constructively interfere: 2 ay = nλ (3) While this relates the difference in path distance to the incident wavelength, this doesn t tell us much about d, the space between atoms x & y. To get the equation in terms of the spacing between the atoms, we notice that d is the hypotenuse of a right triangle with sides ax and ay (or bx and by ). Using trigonometric identities: dsinθ = ay (4) dsinθ' = yb (5) These equations are identical due to elastic scattering, which means θ = θ'. Substituting these expressions into equation (3) yields the Bragg equation (1), which fully describes the conditions necessary for determination of the distances between atoms in a crystal lattice. In fact, this is not limited to just atomic crystals. Any periodic structure will diffract in exactly the same fashion as described in equation (1). This allows for large domains to be imaged (as in our case, 1-6 μm microspheres that we will use to represent atoms in a crystal). The only requirement is that the 2

3 wavelength and the size of the diffraction grating need to be on the same order of magnitude for interference to occur. There are many different d-spacings in a crystal. While difficult to understand, imagine walking through a rock farm (yes, they exist) and the rocks are planted at regular intervals. Look directly down a row, the rocks will obviously be in lines like in Fig. 1, but there are also other angles in which it will appear the rocks are in a periodic structure when you look at it diagonally as shown in Fig. 2. In crystallography, each of the ways that particles line-up is referred to as a lattice plane (also called crystallographic plane). Each plane is capable of scattering light with some efficiency which we detect as a diffraction pattern. Figure 2. Multiple lattice spacings are obtainable through the same crystal structure. As the incident radiation passes through these many planes, the light is scattered in different patterns and intensities. Note how the space between planes (d) decreases as the planes tilt. When we focus light off a single crystal domain, we should see a pattern of bright spots similar to those in Fig. 3 A. Several spots that share the same distance from the center form a group; several groups may be observed in a single diffraction pattern. Each group represents a different d value, or lattice spacing, within a material. However, if we do not focus the light, we no longer see a spot diffraction pattern (rather, rings instead) as shown in Fig. 3B. The reason for the difference in the interferograms is important. In Fig. 3A, the diffracted light interacts with only one crystal domain, and it can only be oriented in one direction. As such, only singular bright spots are observed. However, if the light is diffracting off multiple crystal domains as in Fig. 3B, it is a certainty that each crystal domain will have a different orientation with regard to the incident radiation; this is termed powder crystallography. This means, for each crystal the radiation strikes, the distance from the center to a diffraction spot remains the same but the absolute position of the bright spots (from different crystal domains) would vary across the entire circumference. If a sufficient number of crystal domains are present, the multitude of bright spots forms a ring. 3

4 Miller Indices: We need a system for describing the lattice planes of the sample. Although we can draw the various lattice planes as shown in Fig. 2, all of this illustration work rapidly becomes cumbersome. Fortunately, a mathematical shorthand method exists for labeling the lattice planes. These labels are called Miller indices and can be easily understood once we define the unit cell for our 2- dimensional crystal. The unit cell is the smallest repeating unit that describes the crystal lattice. For our hexagonal crystals of identical microspheres, Figure 3 A. Diffraction spots are obtained when shining light off of a single crystal there are many ways to draw a unit cell; as such, domain. B. Rings are seen when diffracting off of several crystal domains. chemists have defined a conventional unit cell for a given crystal structure. In our case, we are studying 2-dimensional hexagonal lattices; the unit cell conventionally used to describe it is shown in Fig. 4 (left). Now we can define two crystallographic axes, which are the directions that we tile-in the unit cell repeatedly to re-create the whole crystal. As shown below, our crystallographic axes are just sides of the unit cell; as you can see, adding a unit cell at each crystallographic axis tick-mark re-creates the whole structure. The Miller indices are now definable by these tick-marks. Figure 4. Left: The unit cell of a 2-dimensional hexagonal close packed lattice. Note that the crystallographic axes follow the sides of the unit cell. Addition of a unit cell at each tickmark, and further tiling up, re-creates the whole crystal. 4

5 Figure 5. Examples of calculating Miller indices for various patterns of lattice planes. There is always a negative analog to the positive lattice plane. A: (0 1) and (0 1 ). B. (1 0) and (1 0). C. (1 1) and (1 1 ). D. (1 2) and (1 2 ). E. (2 1) and (2 1 ). F. (3 1) and (3 1 ). To determine the Miller index for a set of lattice planes, first refer back to Fig. 2 where we showed how to generate multiple lattice spacings, which are defined by the lattice planes (the lines). The same and more examples are shown in Fig. 5 above. For example A, we drew the most simple lattice planes like in Fig. 1. To determine the Miller index, we look at the same planes using the crystallographic axes; the only rule is to make sure the plane(s) fit within or touch the sides of the unit cell (coordinates on both axes) and that the plane(s) do not cross the center. The lattice planes are represented by two dotted lines. Next, examine where the plane crosses each crystallographic axis. In example A, the plane never crosses the horizontal axis; in this case, we say the line crosses this axis at. The upper plane crosses the vertical axis at 1. The Miller index is then the reciprocal of these intersection coordinates, so the Miller index is (1/ 1/1) or (0 1). Note that it is possible to define two planes in A, as such, the lower dotted line has a Miller index of (1/ 1/-1) or (0 1 ), where the bar indicates a negative sign. Examples B & C are similar to A, but the diffraction occurs in different direction. Example C is a little different in that the two possible lattice planes cross at coordinates 1,1 and - 1,-1, giving (1 1) and (1 1 ) Miller indices. Note the six possible diffraction patterns all have the same d-spacing ( 3/2 microsphere diameter). In example D, the top lattice plane crosses at coordinates 1, 1 2 and the bottom at -1, -1 2 ; these diffractions are in the (1 2) and (1 2 ) directions. 5

6 Note the d-spacing is smaller than before, it s just the radius of the microsphere. Likewise, example F has plane crossings at 1 3, 1 and -1 3, -1; we say that these diffractions occur in the (3 1) and (3 1 ) directions. The (3 1) and (3 1 ) diffractions are due to lattice planes with smaller d- spacing than examples A through D. Check the handout that accompanies this lab for more examples. II. Experimental This lab uses laser light diffraction to study the 2-dimensional crystal of microspheres; this is completely analogous to examining inorganic crystals or DNA with X-ray scattering. Homogeneous microspheres can pack into hexagonal crystal lattices; however, after a certain length, the crystal may turn in another direction and form a new crystal domain as shown below. Regardless, these ordered arrays will create diffraction gratings that produce interferograms similar to that observed in Fig. 3 using a laser as the light source. Ringed diffraction patterns are produced by scattering of multiple domains of microsphere crystals (Fig. 3B) while single crystal domains produce point diffraction patterns as shown in Fig. 3A. To obtain the spot diffraction pattern from a single domain, the laser needs to be focused (the diameter of the incident radiation needs to be narrowed) as the unfocused beam will always interrogate multiple domains. The distance between each microsphere (and thus their size) can be determined by analysis of the diffraction patterns. Arrays of different sized microspheres have already been prepared. For each microsphere solution, the sample was diluted and a single drop was placed on a microscope slide to dry overnight at 4 o C. The slow evaporation allows the microspheres to pack into a highly ordered array from which you will measure several diffraction patterns. You are to: 1. Obtain 3 pre-cast glass cover slips that contain different sizes of microspheres. These will be analyzed for the periodicity of their packing domain and for the average distance between microspheres. Record the identity of each slide on your interferogram. 6

7 2. Turn the laser on. *DO NOT LOOK DIRECTLY AT THE LASER* While the laser will not make you go blind, it can cause pain. Be careful when passing in the plane of the laser (bending down) and do not look at anything at the level of the laser. *Treat the laser with respect* 3. Align the glass slide with the laser beam such that the laser passes perpendicularly through the sample. If the sample is not perpendicularly aligned, the diffraction rings will not be concentric and will be difficult to analyze. Ensure that the interference rings are circular. 4. Put the laser about 12 inches in front of the sample, which is itself also about 7 inches in front of the 90 clipboard. This allows for space to put in the converging lens and further alignment such as moving the sample to obtain a proper focus. Adjust the sample while the laser is incident upon it. As the sample slowly moves in and out of focus, the clarity of the rings will change. Find a position for the sample in which the sharpness of the rings is at a maximum. To ensure the circles are concentric, you can measure the distance between two rings at multiple points. If the distances are equal then the sample doesn t need angular adjustment. Take a photograph as well. 5. Measure the distance from the sample to the center spot and trace the interferogram on a piece of white paper. 6. Next, a lens should be placed between the laser and the sample. This causes the diameter of the laser beam to be narrowed and will allow for imaging of a single crystal domain. Care must be taken to insure the resultant laser beam is still perpendicular to the sample; it may not be if it doesn t go through the center of the lens. Furthermore, the sample must be placed at the lens focal length, which is 6 inches. At the focal distance, the diffraction pattern should appear as discreet bright spots. You can adjust the sample further to make the bright spots as small and spherical as possible. Elongated or oval/rod shaped diffraction spots are not usable for the calculations. This can take some time, so don t get frustrated. Take a photograph of the pattern. 7. Measure the distance from the sample to the center spot and trace the interferogram on a piece of white paper. 8. The interferograms should be similar to Fig. 3A; the bright spots should be uniform. If they are not circular, the sample needs to be adjusted (the angle or distance). If the spots are unfocused, then the sample needs to be moved to obtain sharp spots on the paper on the clipboard. 7

8 9. Repeat this procedure for each microsphere sample, obtaining interferograms from both multiple domains (ring diffraction) and focused single domains (spotted diffraction). These inter-ferograms will be analyzed to calculate the distance between microspheres (and thus their diameter) in each monolayer. 10. Before packing up make sure you a) have single crystal diffraction patterns for each microsphere sample, b) recorded the distance from the sample to the clipboard, and c) turned off the laser and unplugged it. III. Data Analysis The first step to analyzing the diffraction patters is to determine the d-spacing, which is itself a function of the lattice planes (identified by their Miller indices) that caused the diffraction to begin with. Shown below is an example of the geometry of the setup; what is paramount is the determination of the diffraction angles θ n for each ring or spot. This is a fairly trivial task, as you should know R (clipboard-to-sample distance) and the center-to-(ring or spot) distance using a ruler and your traced diffraction pattern. In this example, θ 1 = atan(r 1 /R). Next, you plug in these values into the 2-D Bragg equation: dsinθ = nλ where d is the d-spacing you re looking for; θ was determined above for each ring or spot, λ is 543 ± 1 nm; the diffraction order n will be 1 as the higher order transitions (n = 2, 3 etc.) are very difficult to see. Note the factor of 2 is missing in the 2-D Bragg equation (compared to eq. 1); this is because we are working with one less dimension. 8

9 Now that you have all the d-spacings, you must now calculate the center-to-center distance between micro-spheres; this is also the microsphere diameter. First, re-examine Fig. 5 and note that, no matter what lattice spacing is causing diffraction, the hexagonal packing and microsphere size is nonetheless identical. This means for each d spacing analyzed, the interparticle distance (D) from each microsphere must still be the same. As such, if we can determine a relationship between the d-spacing and the center-to-center microsphere distance D, then every ring or diffraction spot will allow us to calculate D. Unfortunately, the relationship between d (diffraction spacing) and D (microsphere center-to-center distance, or the microsphere diameter) depends on which diffraction ring or spot you re analyzing. For example, let s study diffraction along the (0 1) direction (Fig. 5 A). This is a crystal of spheres overlapped with the (0 1) lattice planes (solid lines). Now, you have to practice geometry; here we draw an equilateral triangle with a length 2 D; each angle is 60 (second step above). Now if the equilateral triangle is bisected as shown in the third step, a triangle is made with a long edge of 2 d and a short edge of D; the relationship between the two is now tan(60 ) = 2 d/d or D = 2 d/ 3. Note that a similar analysis will show that the same relationship exists between d and D for the (0 1 ), (1 0), ( 1 0), (1 1) and ( 1 1 ) diffraction spots as they scatter light by the same angle. Now do you see why Miller indices are useful? Let s look at a smaller d-space, higher angle scattering direction like the (2 1) and (1 2) and figure out the relationship with d and D. In the diagram to the right we can see that in both cases, 2 D = 4 d. No math is required! Same is true for the (2 1 ), (1 2 ), (1 1 ) and ( 1 1) diffraction spots. 9

10 Last, let s examine the (3 1) scattering direction. This is not straightforward; we will begin by drawing a crystal, with the unit cells (red dashed lines) and (3 1) scattering vectors (solid green lines) overlaid. The starting point for the analysis is labeled with an x. We locate a point along the crystallographic axis where a right triangle can be made with the (3 1) scattering vector forming the hypotenuse. This is 2.5 D below the x, and from simple geometric principals, the distance from this new point to the center of the adjacent sphere is 3 D/2; this allows for the determination of all the angles of the right triangle. Next, we extend the line down the crystallographic plane until we can make another right triangle where the crystallographic axis now becomes the hypotenuse. At this point, we can determine that the relationship between D and d is sin(19.1 ) = 3d/3D, which is equivalent to d = 3 D/(2 7). The same relationship exists for many other scattering directions as tabulated below. The relationship in the next, most scattered direction is sin(13.9 ) = 4d/4D, which is equivalent to d = 3 D/(2 13). The main issue to note is that, whether analyzing rings or spots, the following trend exists in 2-dimensional hexagonal close-packed arrays as they spread further and further from the central (undeflected) laser. In terms of increasing angular deflection: Miller Indices Structure factor (1 0) (1 0) (0 1) (0 1 ) (1 1) (1 1 ) D = 2 d/ 3 (1 1) (1 1 ) (2 1) (2 1 ) (1 2) (1 2 ) D = 2 d (1 3) (1 3 ) (1 3/2 ) (2 3 ) (2 3) (1 3/2) (3/2 1 ) (3 2 ) (3 2) (3/2 1) (3 1 ) (3 1) (2 1) (2 1 ) (1 2) (1 2 ) Too many to list! sin(19.1 ) = d/d or D = 2 7 d/ 3 sin(13.9 ) = d/d or D = 2 13 d/ 3 10

11 Note each grouped Miller index in the table scatters light into the same ring ; this is why the D and d relationships are the same. Thus, all you need to do is analyze the closest set of spots to the center (or the smallest ring) and use the relationship D = 2 d/ 3. Take the next closest set of spots (or the next largest ring) and calculate the microsphere diameter via D = 2 d. The next nearest set of spots or next largest ring should have D = 2 7 d/ 3 and so forth according to the table above. It is unlikely you will observe more than 4 sets of spots or more than 4 rings. IV. Lab Report There should be two diffraction patterns obtained per sample (ringed patterns from multidomains and spotted patterns from single domain interferograms). Analyze each one separately to determine the interparticle separation (D) from as many d-spacings as possible; provide the data and results in tabular form as suggested in the handout. Do not forget your error analysis! Estimate the uncertainty of the measurements which you made with a ruler and propagate the error in your reported calculations. Hopefully, all data will yield similar values of D per sample no matter whether a spot or ring is analyzed, or whether from high- or low- angular diffractions. When analyzing the single domain interferogram (the spotted data obtained using a lens), make assignments for the Miller indices for as many of the spots as you can. Obviously, it is almost impossible to differentiate Miller indices with the same d-spacing [such as (0 1), (1 1) etc.]; note this in your write-up and just do the best you can. We have also included several examples of Miller indices, and their d to D relationship, in a handout; this should allow you to identify which spots correspond to which groups of Miller indices. Include a copy of the analyses of the diffraction patterns, as well as photographs and traces of the diffraction patterns in the results or as an appendix; this is up to your discretion. A summary of the findings (the average calculated D per sample with error), should be in the discussion and conclusions. In the discussion, aside from summarizing the results, please comment whether there are significant differences in single vs. multi-domain data and whether one is more accurate (is this supported by error analysis?). Also, are better data obtained with larger or smaller microspheres, and why? What other factors may have influenced the data and subsequent analyses, and how? 11

12 V. Questions 1. Are the particle diameters that you determined using different d-spacings from the same diffraction pattern consistent with one another? Why or why not? If not, is there a systematic trend? Explain. 2. Suppose a student is able to obtain very nice ringed powder diffraction pattern (with an unfocused laser) but cannot resolve the single crystal diffraction pattern. Why would this happen? Would using a different laser help, say, one with a different wavelength? Explain. 3. Powder diffraction is the analysis of a collection of crystal domains while single crystal diffraction specifically interacts with one species. What is the difference in the information content from powder diffraction vs. single crystal diffraction data, and vice versa? 4. These microspheres packed into a hexagonal close packing lattice. What would a single crystal diffraction pattern of a simple cubic system look similar? Please provide a qualitative description. VI. References [1] Stephanie A. Bosse and Nikolaus M. Loening, Using Two-Dimensional Colloidal Crystals to Understand Crystallography J. Chem. Educ., 2008, 85,