Xray Diffraction and Absorption

Xray Diﬀraction and Absorption

Experiment XDA

University of Florida — Department of Physics PHY4803L — Advanced Physics Laboratory

Objective cesses simultaneously produce xrays. First, the deceleration of beam electrons from colli- The diffraction of xrays from single crystals sions with the target produce a broad contin- and powdered crystalline samples will be in- uum of radiation called bremsstrahlung (brak- vestigated along with the emission and absorp- ing radiation) having a short wavelength limit tion of xrays by high Z metals. The diffrac- that arises because the energy of the photon tion patterns show sharp maxima (peaks) at hc/λ can be no larger than the kinetic energy characteristic angles that depend on the wave- of the electron. Second, beam electrons can length of the xrays and the structure of the knock atomic electrons in the target out of in- crystal. A Geiger-Müllertube is used to de- ner shells. When electrons from higher shells tect the intensity of the diffracted xrays versus fall into the vacant inner shells, a series of dis- scattering angle. Crystal properties are deter- crete xrays lines characteristic of the target mined from the angular position and intensity material are emitted. of the peaks. In our machine, which has a copper target, only two emission lines are of appreciable in- References tensity. Copper Kα xrays (λ = 0.1542 nm) Generally look in the QC481 and QD945 sec- are produced when an n = 2 electron makes tions. a transition to a vacancy in the n = 1 shell. A weaker Kβ xray with a shorter wavelength A. H. Compton and S. K. Allison, Xrays in (λ = 0.1392 nm) occurs when the vacancy is Theory and Experiment filled by an n = 3 electron. The reverse process of xray absorption by C. Kittel, Introduction to Solid State Physics an atom also occurs if the xray has either B. D. Cullity, Elements of Xray Diffraction an energy exactly equal to the energy difference between an energy level occupied by an Teltron manual, The Production, Properties, atomic electron and a vacant upper energy and Uses of Xrays level, or an energy sufficient to eject the atomic electron (ionization). For the xray energies Xray Emission and Absorption and metals considered in this experiment, the ionization of a K-shell electron is the domi- When an electron beam of energy around nant mechanism when the xray energy is high 20 kV strikes a metal target, two different pro- enough, which leads to a complete absorption

XDA 1 XDA 2 Advanced Physics Laboratory

JODVVÃGRPH [UD\ÃWXEH

2θθθ θθθ θθθ

d θθθ ÃPPÃVOLW &XÃDQRGH VDPSOH WDEOH FOXWFK d sin θθθ VFULEH SODWH OLQH FU\VWDO *0ÃWXEH θ GHWHFWRU Figure 1: The ray reflected from the second plane ÃPPÃVOLW DUP must travel an extra distance 2d sin θ. ÃPPÃVOLW of the initial xray photon and the ejection of 1 θ an electron—a process known as the photoelec- tric effect). If an xray does not have enough energy to cause a transition or to ionize an Figure 2: The xray diffraction apparatus. atom, the only available energy loss mechanism is Compton scattering. Constructive interference of the reflected waves occurs when this distance is an integral of the wavelength. The Bragg condition for The Xray Diffractometer the angles of the diffraction peaks is thus:

Thus, the spectrum of xrays from an xray nλ = 2d sin θ (1) tube consists of the discrete lines superimposed on the bremsstrahlung continuum. This where n is an integer called the order of diffrac- spectrum can be analyzed in much the same tion. Note also that the lattice planes, i.e., the way that a visible spectrum is analyzed using crystal, must be properly oriented for the re- a grating. Because xrays have much smaller flection to occur. This aspect of xray diffrac- wavelengths than visible light, the grating tion is sometimes used to orient single crystals spacing must be much smaller. A single crys- and determine crystal axes. tal with its regularly spaced, parallel planes Our apparatus is shown schematically in of atoms is often used as a grating for xray Fig. 2. The xrays from the tube are collimated spectroscopy. to a fine beam (thick line) and reflect from the The incident xray wave is reflected spec- crystal placed on the sample table. The de- ularly (mirror-like) as it leaves the crystal tector, a Geiger-Müller(GM) tube, is placed planes, but most of the wave energy continues behind collimating slits on the detector arm through to subsequent planes where additional which can be placed at various scattering an- reflected waves are produced. Then, as shown gles 2θ. In order to obey the Bragg condition, in the ray diagram of Fig. 1 where the plane the crystal must rotate to an angle θ when the spacing is denoted d, the path length differ- detector is at an angle 2θ. This θ : 2θ relation- ence for waves reflected from successive planes ship is maintained by gears under the sample is 2d sin θ. Note that the scattering angle (the table. angle between the original and outgoing rays) Assuming d is known (from tables of crystal is 2θ. spacings), the wavelength λ of xrays detected

February 12, 2007 Xray Diﬀraction and Absorption XDA 3 at a scattering angle 2θ can be obtained from Eq. 1.

Powder Diffraction An ideal crystal is an infinite, 3-dimensional, periodic array of identical structural units. The periodic array is called the lattice. Each y point in this array is called a lattice point. ( y The structural unit—a grouping of atoms or ( molecules—attached to each lattice point is y ( called the basis and is identical in composi- tion, arrangement, and orientation. A crystal thus consists of a basis of atoms at each lattice Figure 3: The simple cubic lattice points (dots) point. This can be expressed: and connecting lines showing the cubic structure. lattice + basis = crystal (2) points at the same positions as those of the The general theoretical treatment for the sc lattice but it has additional lattice points determination of the diffracted xrays—their at the center of each unit square (cube face) angles and intensities—was first derived by defined by the grid. Laue. Starting with Huygens’ principal, it in- volves constructions such as the Fourier trans- A primitive unit cell is a volume contain- form of the crystal’s electron distribution and ing a single lattice point that when suitably develops the concept of the reciprocal lattice. arrayed at each lattice point completely fills The interested student is encouraged to ex- all space. The number of atoms in a primi- plore the Laue treatment. (See Introduction to tive unit cell is thus equal to the number of Solid State Physics, by Charles Kittel.) How- atoms in the basis. (There are many ways of ever, we will only explore crystals that can be choosing a primitive unit cell.) The primitive described with a cubic lattice for which a sim- unit cell for the sc lattice is most conveniently pler treatment is sufficient. taken as a cube of side a0. a0 is called the There are three types of cubic lattices: lattice spacing or lattice constant. the simple cubic (sc), the body-centered cu- The bcc and fcc primitive unit cells (rhom- bic (bcc) and the face-centered cubic (fcc). bohedra) will not be used. Instead, bcc and The simple cubic has lattice points equally fcc crystals will be treated using the sc lat- spaced on a three dimensional Cartesian grid tice and sc unit cell (which would then not be as shown by the dots in Fig. 3. As a viewing primitive). aid the lattice points are connected by lines A crystal is completely described by speci- showing the cubic nature of the lattice. fying the lattice (sc, bcc, or fcc) and the po- The body-centered cubic lattice has lattice sition of each atom in the basis relative to a points at the same positions as those of the single lattice point. For the sc lattice, the basis sc lattice and additional lattice points at the atom positions are specified by their relative center of each unit cube defined by the grid. Cartesian coordinates (u, v, w) within the sc The face-centered cubic lattice also has lattice unit cell. u, v, and w are given in units of the

February 12, 2007 XDA 4 Advanced Physics Laboratory

lattice spacing a0, so that, for example, the turns out that the spacings for all possible lat- body-centered position would be described by tice planes in the sc lattice can be represented the relative coordinates (u, v, w) = ( 1 , 1 , 1 ). by 2 2 2 a When a sc lattice is used to describe a bcc d(hkl) = n√ 0 (3) or fcc crystal, one still specifies the position h2 + k2 + l2 of atoms using relative coordinates (u, v, w). where hkl are integers–called the Miller in- But all atoms within the sc unit cell must be dices and n is an integer. Together with Eq. 1 given—including the extra basis atoms that this leads to the Bragg scattering relation giv- would occur at the body-centered or face- ing the angular position of the Bragg peaks centered positions, respectively. 2a λ = √ 0 sin θ. (4) If the unit cell is taken so that its corners h2 + k2 + l2 are at lattice points, each corner lattice point The Miller indices are used to classify the is shared among the eight cells that meet at possible reflections as shown in Fig. 4. For that corner. In order to avoid having lattice hkl = 100, we speak of the 100 (one-zero-zero) points shared among different unit cells, the reflection, which is from the 100 planes, i.e., corner of the cell is usually given an infinites- the planes along the cube faces. The 200 re- imal backward displacement in all three di- flection, (hkl = 200) is from the 200 planes rections so that of all the eight corner lattice which are the 100 planes and an additional points (0, 0, 0), (0, 0, 1), ..., (1, 1, 1), only the set midway between. (It can be considered the (0, 0, 0) site remains in the cell. Then, for 100 planes with n=2 in Eq. 1.) The 110 reflec- example, the bcc lattice—with lattice points tions are from the 110 planes through opposite at the corners and the body center of the sc edges of the cube, and so on. unit cell—would be described as having lat- Not all reflections are of equal intensity. As tice points only at (0, 0, 0) and ( 1 , 1 , 1 ) within 2 2 2 hkl get large, the density of atoms in each each “pulled-back” unit cell. plane decreases and the corresponding peak Exercise 1 Use a drawing to show that the gets weaker. For plane spacings smaller than fcc lattice—with lattice points at the corners λ/2, the formula gives sin θ > 1, and these and face centers of the sc cell—would be de- reflections cannot occur. Most importantly, scribed as having lattice points at (0, 0, 0); as in the case of the bcc or fcc lattices, or 1 1 1 1 1 1 when there is more than one atom in the ba- ( 2 , 2 , 0); ( 2 , 0, 2 ) and (0, 2 2 ) in the “pulled- back” unit cell. sis, the additional atoms can cause reflections which can contribute either constructively or A simple cubic crystal with a one-atom ba- destructively to the reflection. sis has an infinite number of atomic plane sets, The position of each atom j in the unit though different sets have different spacings. cell can be described by its relative coordi- There is an obvious set of planes separated nates uj, vj, wj. Then, the angular position by a0, passing through opposite faces of the of the Bragg peaks is still described by Eq. 4. unit cell. There is also a set of√ equally spaced However, because of the scattering sites within ◦ planes separated by d = a0/ 2, turned 45 each cell, the intensity of the peak will propor- from the planes along the cube faces and pass- tional to the square of the magnitude of the ing through opposite edges of the cube. There crystal structure factor F (hkl): X should also be constructive interference for re- 2πi(huj +kvj +lwj ) F (hkl) = fje (5) flection angles given by Eq. 1 with this d. It j

February 12, 2007 Xray Diﬀraction and Absorption XDA 5

100 110 210

Figure 4: Top view of three possible sets of planes in the simple cubic lattice with their Miller indices. where the sum extends over all atoms in a sin- with the reﬂections from the Cs planes. This gle unit cell. For certain hkl, this factor may may be veriﬁed from Eq. 5. be zero and the corresponding Bragg peak will be missing. F (100) = fCs − fCl

The atomic form factor fj above depends on While for a 200 reflection the type of atom at the site uj, vj, wj. It also varies with θ. For θ = 0, it is approximately F (200) = fCs + fCl proportional to the number of electrons in the In this case, the pathlength difference for the atom. For larger reflection angles, it decreases Cs planes is 2λ, and for the Cl planes, it is λ. due to interference effects of scattering from Thus, the reflections from each kind of plane different parts of the atom. Determining f are in phase. as a function of θ from measurements of in- In fact, the CsCl structure factor for the tensities in the Bragg peaks is possible, but general (hkl) is difficult because the apparatus collection effi- πi(h+k+l) ciency usually also depends on the scattering F (hkl) = fCs + fCle (6) angle. and gives fCs + fCl when h + k + l is even and Nonetheless, qualitative information is ob- fCs − fCl when h + k + l is odd. Thus, one tainable from the intensity of the Bragg peaks. might expect that superimposed on a gradual Consider the CsCl structure. It has a sim- reduction in peak intensities as the scattering ple cubic lattice structure and a basis of two angle increases, the spectrum would show that atoms. One atom, say Cs, can be considered peaks for which h+k+l is even are larger than at the corners of the sc unit cell (0, 0, 0) and those for which h + k + l is odd. the other, say Cl, will then be at the body Consider next a body-centered cubic crystal 1 1 1 centered positions ( 2 , 2 , 2 ). The 100 reflec- with a single atom basis. Using the sc unit cell tion corresponds to waves reflected from ad- we would have to include one atom at the sc jacent 100 planes of Cs atoms having a path lattice site (0, 0, 0) and an identical atom at length difference of λ. This reflection should 1 1 1 the body-centered site ( 2 , 2 , 2 ). The crystal now be reduced in intensity due to reflections structure factor becomes: from the mid-planes of Cl atoms which would iπ(h+k+l) be in phase with one another but out of phase Fbcc(hkl) = f(1 + e ) (7)

February 12, 2007 XDA 6 Advanced Physics Laboratory

which shows that the crystal structure factor GHWHFWLRQ SODQH for the bcc lattice is non-zero (and equals 2f) LQFLGHQW only if h + k + l is even. Thus peaks with EHDP SRZGHU VFDWWHULQJ h + k + l odd would be entirely missing. VDPSOH FRQH

Exercise 2 Show that the crystal structure factor for a crystal with the fcc lattice type and θ a one atom basis is given by:

iπ(h+k) iπ(k+l) iπ(l+h) Ffcc(hkl) = f(1 + e + e + e ) (8) Then use this result to show that Ffcc is zero Figure 5: A powder sample illuminated by an unless hkl are all even or all odd. xray beam scatters radiation for each Bragg angle into a cone. Next consider the class of crystals having the NaCl structure. NaCl has an fcc lattice Exercise 4 Visualize an NaCl-type crystal structure with a two-atom basis. Putting the with a lattice spacing a0 and explain how, if Na atoms at the normal fcc sites of the sc both atom types are identical, the crystal is cell—the corners and the face centers—the Cl equivalent to a simple cubic crystal with a one atoms will be at the body-centered position atom basis and a lattice spacing a0/2. Now, and at the midpoint of each edge of the cell. show how this works out in the math. As- The relative coordinates in the sc unit cell are sume the atomic form factors in an NaCl-type crystal are exactly equal: fNa = fCl. Use the 1 1 1 1 1 1 Na: 0,0,0; 2 , 2 ,0; 2 ,0, 2 ; 0, 2 , 2 results derived in the previous exercise along with Eq. 4 to show that the xray scattering Cl: 1 ,0,0; 0, 1 ,0; 0,0, 1 ; 1 , 1 , 1 2 2 2 2 2 2 would then be the same as that of a simple cubic crystal with a one atom basis and half the Exercise 3 Evaluate the crystal structure original lattice spacing. Keep in mind that for factor for the NaCl crystal. Show that F is a simple cubic crystal with a one atom basis, still zero unless hkl are all even or all odd and the crystal structure factor is just the atomic give the structure factor in terms of the atomic form factor for that one atom and all hkl in form factors fCl and fNa. Eq. 4 lead to non-zero scattering. (Don’t be concerned about a factor of 8 difference in the While there are no simple cubic crystals structure factors. It is just the ratio of the with a one-atom basis in nature, the potas- number of atoms in the unit cells for the two sium chloride (KCl) crystal behaves nearly cases.) so. Potassium chloride has the NaCl structure with the potassium having atomic num- For single crystal diffraction, the crystal ber 19 and the chlorine having atomic number must be properly oriented to see reflections 17. However, because of the ionic bonding, from particular lattice planes at particular an- each atomic site has roughly the same num- gles. With a powder sample, all crystal ori- ber of electrons (18) and scatter xrays nearly entations are present simultaneously and the equally well, i.e., have nearly the same atomic outgoing scattered radiation for each Bragg form factors. diffraction angle forms a cone centered about

February 12, 2007 Xray Diffraction and Absorption XDA 7 the direction of the incident radiation as a vertical edge of the sample post. Use shown in Fig. 5. As the detector slit passes the 3 mm slit in slot 13 on the detector through a particular cone, the GM tube be- arm bench, and the 1 mm slit in slot 18. hind the slit detects an increase in the radiation and a peak in the spectrum will be 3. Place the LiF single crystal with the flat recorded. matte side centered against the sample post (see section D27.30 of the Teltron manual). Procedure 4. Turn on the keyed switch and move the Setup automatic shutoff timer off 0. The fil- 1. Open the shield and completely unscrew ament should light but the high volt- the plastic posts holding the motor gear age will remain off and xrays will not be ◦ against the detector arm gear. Move the present. Set the arm back to 0 and sight motor completely out of the way of the through the detector slits, past the crys- inner angular scale around the sample ta- tal sample and through the primary col- ble. The motor/detector arm gears sim- limator. All edges should be vertical and ply drive the detector arm. Gears under you should see a reflection of the filament the sample table are responsible for main- light from the slanted surface of the cop- taining the θ : 2θ relationship but they per anode just skimming past the crystal must be set properly at one angle first. face. First, rotate the arm around by hand and 5. Put the GM tube (with the connector note how the sample post rotates half as ◦ wire pointing up) in slot 26. Make sure much. Then, set the arm at precisely 0 the GM tube BNC cable is connected to and check that the scribe marks on op- the top BNC output on the TEL 2807 posite sides of the sample table line up ◦ Ratemeter. This output will have at 0 on the inside θ scale. The raised, dangerously high voltage. Make sure chamfered, semicircular post must be on nothing but the GM tube is connected the side away from the motor. The sam- to this output. Push the CHANNEL SE- ple table can be rotated independently LECT button to monitor channel 3—G/M of the detector arm if the knurled clutch TUBE voltage. Set it to about 420 V. plate is first loosened sufficiently. If both scribe marks cannot be made to line up 6. Check that the high voltage selector exactly at the zeros of the θ scale, they switch on the scattering table (under the should be made to give equal angles on motor) is set for 20 kV. the same side of a line connecting the zeros. Carefully retighten the clutch plate. 7. Move the detector arm out of the way Finger-tighten only! Do not use pliers or of the locking mechanism and close the wrenches. shield. The shield enters the interlock off- center, and must be clicked into a central 2. Place the primary collimator (button position before the xrays will come on. with 1 mm slit) on the exit port of the Xray emissions begin when you depress glass dome surrounding the xray tube. the red XRAY ON button to turn on the Orient the slit vertically by sighting past high voltage supply. If the xrays do not

February 12, 2007 XDA 8 Advanced Physics Laboratory

come on, the shield may need some jig- the θ−2θ table is set properly, then check gling to get the interlocks engaged. If the with an instructor. machine crackles, leave on the filament, but turn off the high voltage by bumping 11. Open the shield and tighten the motor the shield. Wait about 5 minutes for the mounting against the detector arm gear. things to dry out before trying again. The apparatus is now ready to take spectra. 8. Slide the arm around by hand over the full range of angles from about 13◦ to 120◦. 12. Click on the Initialize Angle button. This There has been some trouble with the arm brings up a new window to initialize the rubbing into the shield and getting stuck program so it knows the angle of the spec- during a computerized run. Play around trometer arm. Click as necessary on the with the positioning of the shield so that Step button while checking the angle on it interlocks properly allowing the xrays the spectrometer to bring the arm to some to come on, but such that it does not in- degree marking on the scale imprinted on terfere with the free motion of the rotat- the spectrometer. Enter the angle for this ing arm. marking in the Present Angle control and then click on the Done button. 9. To measure the xray tube current, in- stall the xray tube current cable, plug- 13. Set the Angle control to bring the arm to ging one end into the jack on the base of 120◦ and then back down to 15◦. Then the spectrometer and the other end into try this again. If there is any problem the Keithley model 175 multimeter (set with the free and accurate movement of to 200 µA DC scale). Too high a tube the spectrometer arm using the stepper current will load down the high voltage motor, fix it or notify the instructor be- power supply, reducing the voltage before continuing. low 20 kV. The manual warns not to go above 80 µA. Set it to around 70 µA us- Xray spectrum ing a small screwdriver on the adjustment at the base of the spectrometer next to 14. Click on the Collect Spectrum to obtain XRAY ON button. the spectrum of xrays emitted from the tube. You will then be asked to enter 10. Activate the XRAY data acquisition pro- a file name in which to store the data. gram. If it is not running, click on the Remember to store all data in the My LabVIEW run button in the tool bar. Documents area of the disk. The data is Click on the Monitor Counts button. This only saved at the end of the run (or after displays counts collected in a fixed time aborting). If you hit the stop button interval as chosen by the control just un- in the tool bar while taking data, der the button. Slide the arm around to that data will be lost forever. Use 45−46◦ where there is a strong diffraction the Abort button under the Collect Spec- peak. Find the angle where the count rate trum button to stop a run prematurely. maximizes. It should be within about 1◦ Use the LiF single crystal on the sample of 45◦ and the count rate should be above post and scan over the full range of scat- 1500/s. If it is below this, recheck that tering angles (15-120◦) with a step size of

February 12, 2007 Xray Diﬀraction and Absorption XDA 9

1/8◦. Because the single crystal scatter- following day to turn off the xray machine ing is strong, you can set the Time per and analyze the data. Increase or decrease channel short, around 5-10 seconds and the time per channel to fully utilize the time the signal to noise will be quite reason- the xray machine is on. For example, if you able. Be sure to start at 2θ = 15◦ to get start the run at 10:00 am and will not be back data at the bremsstrahlung cutoff wave- until 3:00 pm the next day, then adjust the length. time per channel for a total running time of 28-29 hours. There is no point in having the C.Q. 1 Explain the spectrum you see. Use xray tube running with the machine not taking Eq. 1 to convert the angles of the various data. No run should be made longer than 30 spectral features to wavelengths. The crys- hours or so, without first discussing it with the tal face is along the 100 sc lattice plane and, instructor. The clamp on the shut-off timer for LiF, a0 = 0.403 nm. Because LiF is an knob is needed to keep the xrays from shut- fcc crystal, there are two planes of atoms ting off after 55 minutes. per (simple cubic) lattice spacing a0 and thus the appropriate plane spacing d for 15. Replace the LiF crystal with a powder use in Eq. 1 is given by d = a0/2. Obtain sample of LiF. Take an overnight run. best estimates of the xray Kα and Kβ wavelengths as well as the low wavelength cut- 16. Repeat with the KCl powder sample. off λc. Compare the line wavelengths with reference values and λc with expectations CHECKPOINT: C.Q. 1 should be based on the tube voltage. answered. Spectra for all crystal and powder samples should be acquired and Powder Diffraction the angles for all peaks in each spectrum should be measured and tabulated. Next, you will take some spectra using powder (microcrystalline) samples. The scattering for powder samples is much weaker than for single Xray absorption crystals because only a fraction of the crystals are in the proper orientation for any particu- In this last investigation, you will study the lar scattering angle. Thus, for powders, you absorption properties of several metallic ele- will need to run overnight scans. You might ments on xrays emitted by the xray tube. The want to take a one- or two-hour spectrum in diffractometer will be used to vary the xray en- class, setting the starting and stopping angles ergy incident on a metal foil placed on the de- to scan over a small range of about 5◦ around tector arm in front of the Geiger-Müllertube. the strongest expected peak to check the sig- The ratio of the count rate taken with the foil nal (peak height) to noise (background height) to the count rate taken without the foil (i.e., ratio for this peak. If the signal to noise is the transmission fraction or transmittance of reasonable, then set it for an overnight scan. the foil) will be determined point-by-point at You should be able to get a reasonable signal each scattering angle 2θ. Bragg’s law will then to noise ratio at 100 seconds per channel. A be used to determine the xray wavelengths full scan from 2θ = 15 to 120◦ in 1/8◦ steps where absorption features are found to occur. will then take about 24 hours. Thus, start a spectrum before leaving and come back the 17. Use the LiF crystal on the scattering post.

February 12, 2007 XDA 10 Advanced Physics Laboratory

18. Take one scan without any filter in place spectrum, the transmittance ratio would then for the 2θ region from 30◦ < 2θ < 55◦ appear (erroneously) to fall at that angle. As with a 1/4◦ step size and an acquisi- another effect, small random variations in the tion time of 10-20 s per point. This will tube current would cause small random varia- be one of the reference spectra. Take a tions in the measured xray intensity and any second reference spectrum, again with- measured xray count might show random vari- out any absorber, for the 2θ region from ations above and beyond what would be pre- 45◦ < 2θ < 75◦ again with a 1/4◦ step dicted by the “square-root” statistics appro- size and about 30-60 s per point. (At the priate for a source of perfectly uniform inten- larger angles the source xray intensity is sity. While a jump in the tube current or weaker and longer acquisition times are an unsteady tube current has been observed needed.) with our apparatus, neither issue typically af- fects the analysis to any significant degree. 19. Next take sample spectra with at least Nonetheless, the principal—that the xray in- four, but preferably all eight foil samples tensity is proportional to the tube current—is available: Zn, Cu, Ni, Co, Fe, Mn, Cr, worth keeping in mind. and V, inserted (one at a time) into the detector arm in front of the GM tube. For If the tube current were perfectly stable, any of the first four use the same settings the spectrum with the absorber should always as for the first reference spectrum and for have a lower count rate than the reference the last four use the settings for the sec- spectrum and the transmittance should be be- ond reference spectrum. It is best to take low unity at all angles. An absorption edge all reference and sample spectra without due to the foil should appear at some scatter- turning off the data acquisition program ing angle above which the transmittance ratio or re-calibrating the angle. This way, all is just a little under unity (indicating a small spectra should have the least amount of amount of absorption) and below which the angular shift between them. ratio is significantly smaller than unity (indicating a larger absorption of the xrays). How- 20. Plot the transmittance of each foil ver- ever, do not be surprised if the region near the sus 2θ by dividing the appropriate spectra Kα and Kβ peaks shows transmission fractions point-by-point. The transmittance at any both much larger and much smaller than one. angle is the ratio of the spectrum with the This will happen if the scattering angles for absorber to the reference spectrum with- the reference and sample spectra are slightly out the absorber. shifted in angle from one another. In the re- gions very close to the peaks where the count The xray source intensity varies with en- rate is changing rapidly, the shift leads to both ergy and that is why a reference spectrum is incorrectly large and incorrectly small trans- taken and the amount of absorption in the mittance ratios. Taking reference and sample sample spectrum is determined by dividing the spectra without shutting down the data ac- two spectra angle by angle. However, keep in quisition program should help minimize any mind that if the tube current varies, there will angular shifts. If the affect of the offset is ob- be proportional variations in the xray inten- served, try to ignore the errors in the trans- sity. For example, if the tube current were mittance near the peaks and concentrate on to fall at some angle while taking a sample finding the absorption edges.

February 12, 2007 Xray Diﬀraction and Absorption XDA 11

C.Q. 2 Read up to learn about the issues weaker one from 0.138 nm xrays; with the involved in xray absorption and answer the same hkl both should give the same a0. If following questions. Explain why the ab- the scattering is too weak, the peak due sorption edge occurs; why absorption at to the weaker 0.138 nm xrays will become angles below the edge suddenly increases. diﬃcult or impossible to detect and only How does the absorption depend on the the peak due to the 0.154 nm xrays will be xray energy? What electronic transitions observed. You must also keep in mind how are involved in the absorption? What hap- the atomic form factor and crystal structure pens if the xray energy is less than the tran- factor are expected to behave. For example, sition energy? Which foils absorb the cop- if the crystal is fcc, then only scattering per Kα line? Which absorb the Kβ line? with all even or all odd values of hkl will Why doesn’t copper self-absorb its own Kα occur. If it is bcc then only scattering with or Kβ emissions? Determine the xray wave- an even hkl sum can occur. Furthermore, length at the observed absorption edge for the atomic form factors decrease with in- each foil and use this to determine the xray creasing scattering angle. Then, since the energy there, say in keV. Plot the absorp- scattering angle increases as hkl increases, the tion edge energy vs. the atomic number Z strongest peaks always have the smallest hkl of the foil element. How are these ener- consistent with the crystal structure factor. gies expected to depend on atomic number? C.Q. 3 Explain why it should be impos- Do an analysis to demonstrate or verify this sible to observe a given hkl peak for the prediction. 0.138 nm xrays without also detecting the same hkl peak for the 0.154 nm xrays. With a crystal having the NaCl structure, Analysis if the last observed peak corresponds to Determining the best crystal lattice constant λ = 0.154 nm and hkl = 331, which other hkl peaks (at lower angles) should also be a0 from the xray spectrum is not trivial. First, use the LabVIEW xray program and cursor to observable for this λ? determine the center of each scattering peak. This is the scattering angle 2θ from which the 21. Prepare a table for each powder sam- Bragg angle θ is then obtained. For each peak ple, with a row for each Bragg peak and you must then assign (make an educated guess columns for: as to) a value for λ (either 0.154 or 0.138 nm) (a) the scattering angle 2θ. and a set of values for (hkl). Using these in Eq. 4 produces a value for a0. Of course, if ei- (b) the true angle θ. ther the λ or hkl are not the actual values for (c) the appropriate xray wavelength the scattering peak, the a0 will not be correct. (0.154 or 0.138 nm). In fact, you must try to hit upon logical com- (d) the values for h, k and l. binations that produce, within experimental error, the same lattice constant for each peak. (e) the calculated lattice constant a0. What does logical mean? For example, if the scattering for a given hkl is strong C.Q. 4 For each sample, print out the xray enough, two peaks should be observed—a spectrum, labeling the Miller indices hkl stronger peak from 0.154 nm xrays and a and the value of a0 for each peak.

February 12, 2007 XDA 12 Advanced Physics Laboratory

With the correct assignments of λ and hkl, and a0 is the true lattice constant. the values for a0 determined from each peak Make spreadsheet columns for am, g1, and g2 should be close, but not identical. There are with values for g1 and g2 in adjacent columns. random errors associated with the determina- Perform a multiple linear regression to Eq. 11 tion of the scattering angle and systematic er- selecting the am values for the Input Y Range rors inherent in xray diffraction work. Sys- (y-variable) and selecting both the g1 and g2 tematic errors arise from xray absorption and columns for the Input X Range (x-variables). from various kinds of misalignments in the ap- The fitted Constant is the best estimate of a0 paratus. They are typically smaller at larger and the fitted X Variable (1 and 2) are the scattering angles and thus better values for best estimates of c1 and c2. Add a column the lattice constant are obtained at higher an- for am according to the fitting formula using gles. Careful modeling of possible sources of the regression values for a0, c1 and c2. Make systematic error (Cullity, Elements of Xray a graph with plots of the measured am (mark- Diffraction, Ch. 11) leads to several predic- ers, no line) and fitted am (smooth line, no tions for systematic effects. markers) vs. 2θ. Both g1 and g2 go to zero for ◦ One prediction is that if the xray beam hits backscattering (2θ = 180 ) and thus a0 may the sample either to the left or right of the be considered the extrapolated fitted value at centerline there will be an error ∆a in the cal- this angle. culated lattice constant given by The general trend of the fitted curve away from horizontality (c = c = 0) indicates the ∆a D cos2 θ 1 2 = − (9) size of the systematic errors while the scatter a R sin θ 0 of the data about the fit (the rms deviation) where R is roughly the distance from the represents the size of the random errors. This source to the sample and D is the xray beam suggests two ways to judge the significance of displacement—positive if the beam is dis- the systematic errors. (1) Determine whether placed toward the detector side. or not c1 and c2 are consistent with zero (in Another prediction is that scattering from comparison with their 68 or 95% confidence atoms deeper into the sample will produce a intervals) and discuss if the g1 or g2 system- lattice error atic errors are present in the apparatus or too small to be observable. (2) Look at the differ- ∆a = κ cos2 θ (10) ence between the rms deviation of the fit and a0 a direct evaluation of the standard deviation where κ depends on the thickness of the sam- of the measured am (equivalent to the rms de- ple and its xray absorption coefficient. viation of a fit with c1 = c2 = 0). Report on these two deviations and discuss how they Thus, the measured values of a0 (let’s now demonstrate the significance of systematic er- call them am) are predicted to be rors.

am = a0 + c1g1 + c2g2 (11) If you found systematic errors are signiﬁ- cant, you might then look to see if your data where can distinguish between g1- and g2-type errors. cos2 θ g1(θ) = (12) The g1 and g2 functions are relatively similar sin θ in that both monotonically decrease with an- 2 g2(θ) = cos θ (13) gle. They diﬀer in that g1 has a larger range

February 12, 2007 Xray Diffraction and Absorption XDA 13 of values. Try fitting with only one and then tor if you would like to explore other xray ex- only the other as the x-variable and report on periments. the differing values for the lattice constant and the rms deviation of the fit. What conclusions can you draw regarding the systematic errors based on the three ways you used to include them? There may also be a systematic error in the measured scattering angles. The gears on the motor and rotating table would dictate that the detector arm should move exactly 1/16◦ per pulse sent to the motor. After a long scan there may be a difference between the angle as determined by the scale on the xray machine and the computer value. While there might be a problem with the stepper motor and gears, e.g., slippage, the cause of the discrepancy may be as simple as inaccuracies in the silk screened angular scale imprinted on the spectrometer. You should investigate this issue and how it might affect your results. C.Q. 5 For each sample, explain the values of hkl observed, give a single best estimate of the lattice constant and uncer- tainty and compare with reference values. Discuss the observed intensity pattern of the peaks with respect to the calculations of the crystal structure factor. Include error estimates where appropriate and tell in the report how you determined them. Also report on the consistency (or lack of consistency) in the systematic error parameters c1 and c2 for different samples. Use a rep- resentative c1 value to get a rough estimate of the beam displacement D. Is this D reasonable?

There are many other experiments that can be performed with this apparatus. The GM tube response can be studied as a function of the operating voltage. The absorption coeﬃ- cient for aluminum can be determined. Con- sult the references or discuss with the instruc-

February 12, 2007