Exercises and Solutions to Chapter 3 of

2 downloads 0 Views 172KB Size Report
Modelling a nickel atom by a sphere the atomic radius r and diameter d become r = (V. 3 ... 3 πr3 \ = 3. 4π ( D r \3. Ni atoms in a cube with edge D, but there are only. 3. 4π /D − .... packing of spheres of volume (4/3)π(10−5mm)3 = 4.189 · 10−15mm3 the number of spheres .... 〈L3〉 = (125 [nm3]+4 · 1000 [nm3])/5 = 825 [nm3],.
A. v1

Exercises and Solutions to Chapter 3 of ”Thin Film Analysis by X-ray Scattering” Wiley-VCH, Weinheim, 2006 Mario Birkholz Thin Film Analysis and Technology Postfach 120610, D - 10596 Berlin

Abstract Chapter 3 introduces in line profile analysis (LPA). For this purpose, standard line profiles are presented and the effect of size- and strain-broadening is outlined. The instrumental box is dedicated to numerical techniques, while the structure box of this chapter deals with crystalline lattice faults. Also an errata list is given here for chapter 3.

Number of crystallite surface atoms

Nickel crystallizes in the fcc structure, having a = 0.3524 [nm]. Assume that a polycrystalline Ni sample comprises of monodisperse cube-shaped grains. Calculate the number of atoms Nsur that reside on the surface versus the total number Ntot within the cube for a grain edge of D = 2 [µm]. Depict Nsur /Ntot for D between 2 [nm] and 2 [µm]. Solution There are 4 atoms per unit cell, i.e. the atomic volume is VΩ = 0.35243 /4 = 1. 094 × 10−2 [nm3 ]. Modelling a nickel atom by a sphere the atomic radius rΩ and diameter dΩ become     3 1/3 3 1/3 rΩ = VΩ = 1. 094 × 10−2 = 0.1377 [nm] and dΩ = 0.2754 [nm]. 4π 4π    3 4 3 D There are Ntot = D3 / πrΩ3 = Ni atoms in a cube with edge D, 3 4π rΩ  3 3 D − 2dΩ within the cube volume. but there are only 4π rΩ  3  3 D − (D − 2dΩ )3 It is concluded that Nsur = and thus 4π rΩ3 D3 − (D − 2dΩ )3 (D − 2dΩ )3 d Nsur = =1− = 1 − (1 − 2 )3 . Ntot D3 D3  D  3 20003 − (2000 − 2 · 0.2754)3 For D = 2 [µm] it holds that Nsur = 4π 0.13773  3 3 2000 = 6. 04 × 108 and Ntot = = 7. 31 × 1011 . 4π 0.1377

0.1

0.01

0.001 200 400 600 800 1000 1200 1400 1600 1800 2000 x

Logarithmic scale plot of 1 − (1 − 2

0.2574 3 ) D

Whereas the number of surface atoms is in the range of 10% for grains with D in the nanometer range, their relative number declines below 1/1000 for D = 2 [µm]. This is the reason, why some researchers argue that ”nanotechnology brings the surface into the bulk”. 1

2

B.

Pearson function

D.

Compare different Pearson functions with varying exponent m, but constant parameters I0 and 2w. How will the tails of a peak evolve for increasing m? Solution The tails of a Pearson function will diminish for increasing m.

Asymmetric profiles

Identify the range of FWHM 2w and µt products, for which the absorption factor Aθ2θ may account for line profile asymmetries above the 1%, 5% and 10% level. Solution 2µt )). If θ1 and θ2 are the θ coordinates The absorption factor is Aθ2θ = (1−exp(− sin θ at half maximum on the left and on the right side of the reflection (2w = θ2 − θ1 ), the asymmetry fa can be considered to be the difference of absorption factors at θ2 and θ1 normalized to the absorption factor at θ1 . This would read exp(−2 sin(θµt+2w) )−exp(−2 sinµtθ ) Aθ 2θ − Aθ1 2θ1 1 1 = fa = 2 2 1−exp(−2 sinµtθ ) Aθ1 2θ1 1 2µt  − θ1 From this result the equation 2w = − arcsin  2µt 2µt − − ln fa −fa e

sin θ1

+e

sin θ1

may be derived relating 2w and µt for a given asymmetry fa and diffraction angle θ1 . According to it the figures were calculated that display the course of fa = 0.01, 0.05 and 0.1 levels for θ1 = 22.5◦ and 45◦ in the (µt, 2w) plane.

C.

Background

List the experimental causes that may contribute to the background in a θ/2θ pattern. Solution The background in a θ/2θ pattern can be caused by a) x-ray fluorescence radiation of the sample b) scattering by air c) scattering by the aperture edges d) scattering by the substrate e) the bremsstrahlung part of the spectrum of the primary x-rays. Those parts of the background that are due to a) and e) may principially be reduced by improving the energy resolution ∆E/E of the detector. However, this may often not easily be performed and may also be associated with a loss of intensity at E0 . The equilibrium between signal and background is a sensitive trade off and should always be checked carefully before starting a measurement.

3

4

E.

Rachinger correction

G.

Sketch the course of a computer program to perform the Rachinger correction for a set of measured (2θi , Ii ) data. How would you handle the intensity correction if the 2θ coordinate according to Eq. (3.12) does not belong to the set of measured data? Solution 1. Read all 2θi and Ii1+2 , i = 0...n − 1 2. Determine ∆2θ ⇐ 2θi − 2θi−1 . For equidistant sampling 2θi = 2θ0 + i∆2θ 1+2 1 3. Set In−1 ⇐ In−1 4. Loop j = n − 2 down to 0 { 2θ ⇐ 2θ0 + j∆2θ  sin θ) − 2θ0 /∆2θ, rounded to next integer i ⇐ 2θ + 2 arcsin( λλα2 α1

if i < n − 2 then Ij1 ⇐ Ij1+2 − R21 Ii1 } If there are no 2θi they can be aproximated from the indices of the reflection and its estimated line width.

F.

Scherrer factor

Show that the integral breadth (3.26) converges versus 0.903N when the integration limits are set to the first minima ±1/N and the number of unit cell N diverges to infinity. Would it make sense to define the Scherrer constant KS = 0.903 for cube-shaped crystallites? Solution 1/N 2 πNξ In the integral −1/N sin dξ the integration limits go to zero for N → ∞. sin2 πξ Therefore the integrand can be simplified to Iint = Because the integrand is symmetric one can write

Iint =

2 π2

limN→∞

1/N 0

sin2 πNξ dξ ξ2

=

2 π2

limN→∞



1/N

2 πN ξ limN→∞ −1/N sin(πξ) 2 dξ.

cos 2πNξ 2ξ

+

 2πNξ Si(2πNξ)−1 1/N 2ξ 0

where Si(x) is the sine integral. Thus

2πNξ 2π − N2 + 2πN Si(2π) − limξ→0 cos 2ξ − Iint = π22 limN→∞ N cos 2 2

1 2ξ

+



2πNξ Si(2πN ξ) 2ξ



2πNξ for small ξ: cos 2ξ ≈ 2ξ1 and Iint = π22 limN→∞ {πN Si(2π) − limξ→0 πN Si(2πN ξ)} Since Si(0) = 0 and Si (2π) = 1. 41815 1576 we arrive at Iint = π2 N Si(2π) = 0. 90282 N and βQ = 0. 90282 D2π cub Normally, the central peak overlaps with the satellite peaks in the peak line profile because (i) the crystallites are too large and therefore the distance between the satellite peaks and the central peak is small and (ii) - most significant - the distribution of crystallites often is far from monodisperse. Therefore it is more useful to set the integration limits to ±1/2.

L space resolution

Assume that a Bragg peak is to be analyzed that was measured at 33◦ with λ = 0.154 [nm]. Calculate the resolution ∆L in L space when the 2θ data window ranging over (i) 32 − 34◦ and (ii) 30 − 36◦ . What starting and end values 2θS and 2θE have to be chosen if a resolution of 0.23 [nm] is desired in L space? Solution  0.154/2   (i)  = 4. 60 [nm]  λ/2 sin(π34/2/180) − sin(π32/2/180) ∆L = = 0.154/2 sin θE − sin θS   (ii) = 1. 53 [nm]  sin(π36/2/180) − sin(π30/2/180) It can be concluded that a resolution ∆L of less than 1 [nm] is difficult to achieve by usage of CuKα, since this would require the availability of a broad single peak region in the diffractiogram free of any other reflections. In particular, a resolution ∆L comparable to the atomic size would be difficult to achieve. Assuming the peak to occur at 2θ = 33◦ an intended resolution of the Nickel atom, ∆L = 0.23 [nm], would require that no peaks occur between 2θS = 33.0◦ − 20.11◦ = 12.89◦ and 2θE = 33.0◦ + 20.11◦ = 53.11◦ , since 0.154/2 = 0.23 has the solution x = 20. 11◦ sin(π(33 + x)/2/180) − sin(π(33 − x)/2/180) It is evident that such an L space precision appears purely hypothetical, which holds in particular for thin film work.

H.

Fourier transforms of standard functions

What are the discrete Fourier transforms H(L) = F (h(Q)), when h(Q) is identified with the Gauss, Cauchy and the Voigt function? Make use of the integral breadth representation of the model functions Solution Because all three model functions are symmetric, only the cosine term is needed. According eq. (3.20) the discrete Fourier transforms are: Gauss: hG (Q) = exp[−(Q − Q0 )2 π/βG2 ]  2 2 HGn = N−1 j=0 exp[−(Qj − Q0 ) π/βG ] cos(2πnj/N ) N−1 2 ) HG (L) = j=0 exp[−(Qj − Q0 ) π/βG2 ] cos((QE − QS ) Lj N For N → ∞ the Fourier transform HG (L) becomes a Gauss function, too. Cauchy: hC (Q) = [1 + (Q − Q0 )2 π 2 /βC2 ]−1  2 2 2 −1 HCn = N−1 cos(2πnj/N) j=0 [1 + (Qj − Q0 ) π /βC ] N−1 2 2 HC (L) = j=0 [1 + (Qj − Q0 ) π /βC2 ]−1 cos((QE − QS ) Lj ) N ∞ Voigt: hV (Q) = βCββG −∞ hC (ξ)hG (Q − ξ)dξ HV n =

N−1

HV (L) = 5



∞ β −∞ hC (ξ)hG (Qj − βC βG  N−1 ∞ β −∞ hC (ξ)hG (Qj j=0 βC βG

j=0

6



ξ)dξ · cos(2πnj/N ) 



− ξ)dξ · cos((QE − QS ) Lj ) N



I.

Number of crystallites

J.

A polycrystalline specimen comprises of spherical grains and exhibits a monomodal distribution with a diameter of 10 [nm]. How many crystallites cover a surface of 1 [cm2 ] when the film thickness is 1 [µm]? Assume the orientation of crystallites is isotropic and calculate how many grains would contribute to a Bragg reflection when the divergence slit is set to 0.25◦ and 1◦ ? How are the results affected when the diameter expands to 50 and 100 [nm]? Solution The volume of a 1 [cm2 ] film of thickness 1 [µm] is 0.1 mm3 . Assuming a close packing of spheres of volume (4/3)π(10−5 mm)3 = 4.189 · 10−15 mm3 the number of spheres is 0.740 · 0.1/4.19 · 10−15 = 1.77 · 1013 . When the diameters of the crystallites 13 13 = 1. 416 × 1011 or 1.77·10 = were 50 or 100 nm, the sample would contain 1.77·10 53 103 10 1. 77 × 10 spheres. Only those crystal lattice planes scatter into the detector, for which the scattering vector Q is oriented perpendicular upon them. The fraction of all such oriented crystallites compared to the full set is given by the ratio δ δa /2 p /2 1  dϕ sin ϑdϑ 2π 0

−δa /2

with δa and δp accounting for the axial and parallel divergence of the x-ray beam (Instrumental Box 3). The first follows from δa /2 = arctan Ba /2/R with Ba = 10 [mm], while the latter is the divergence defined by divergence slit, δp = δds . δa (1 − cos δds /2). The integral is solved to yield 2π For δds = 0.25◦ it holds (1 − cos δds /2) = 2.38 · 10−6 , for δ = 1◦ the factor is 38.1 · 10−6 . For an assumed diffractometer radius R = 200 [mm] it is obtained δa = (π)−1 arctan(10/2/200) = 7. 9561 × 10−3 , i.e. 2π δa (1 − cos δds /2) = 1. 8936 × 10−8 and 3. 0313 × 10−7 2π for δds = 0.25◦ and 1◦ , respectively. The absolute numbers of contributing crystallites for different diameters D and divergences δ are given in the following table δds = 1◦ D δds = 0.25◦ 3 10 [nm] 335.2 × 10 5365 × 103 50 [nm] 2.681 × 103 42.92 × 103 3 100 [nm] 0.335 × 10 5.365 × 103 For highly symmetric lattices the number is enhanced by a larger than unity multiplicity factor (m > 1). 7

Interference function 1

Derive the interference function ℑ(Q) for cube-shaped crystallites (Eq. (3.25)) from the general-shape interference function (Eq. (3.29)) Solution sin2 (aP (L)Q/2) Eq. (3.29) reads ℑ(Q) = dxdy/ dxdy. sin2 (aQ/2) For cube shaped crystallites P (L) becomes Pcub (L) = N3 . sin2 (aN3 Q/2) dxdy/ dxdy. Accordingly, ℑcub (Q) = sin2 (aQ/2) dxdy runs over the projected area of the crystallite onto the (x, y) plane. 2 3 Q/2) Because sinsin(aN is independent of x and y, the ratio can be written as 2 (aQ/2) sin2 (aN3 Q/2) sin2 (aQ/2)

K.



dxdy/



dxdy =

sin2 (aN3 Q/2) sin2 (aQ/2)

= ℑcub (Q) , which is Eq. (3.25).

Interference function 2

A polycrystalline thin film may be composed from a set of elliptically shaped crystallites with major axes a, b and c. The variation of each axis may be accounted for by a log-norm distribution. Calculate the intensity / interference function ℑ(Q) for a Bragg reflection hkl. Solution The solution has to make use of Eqs. (3.38) and (3.29). Assuming the c axis of each ellipsoid to be oriented normal to the film surface, the column height distribution is   2

2

2

2

1 − xa2 − yb2 P (L(x, y)) = 2d c2 (1 − xa2 − yb2 ) = 2c d In this formula the crystal structure is assumed to be cubic with the cell constant d (not a, to avoid confusion with the major axis a). Insertion into Eq. (3.29) yields ℑ′ell (Q) =

1 πab



sin2 (d· 2c d

2

2

1− x2 − y2 ·Q/2) a

b

sin2 (d·Q/2)

ellipse

1 πab sin2 (d·Q/2)







2

dxdy

y2 b2

x2 a2

= − · Q) · dxdy ellipse sin (c 1 − Herein, πab = ellipse dxdy is the area of the ellipse with major axes a and b. When each axis has a log-normal distribution interference function Eq. (3.38) becomes for the special case considered here   ℑell (Q) =



gell (a,b,c) πab sin2 (d·Q/2)



ellipse



sin2 (c 1 −

x2 a2



y2 b2

· Q) · dxdy dadbdc

with the log-normal distribution of the crystallite shapes 3

gell (a, b, c) =

(2π)− 2 abc ln γa ln γb ln γc

2

(ln b−ln b0 )2 c−ln c0 )2 − (ln2(ln ) 2(ln γb )2 γc )2 √ 2 2 2 replaced by 2π h d+k +l .

a−ln a0 ) · exp(− (ln2(ln − γa )2

For a cubic crystal structure Q may be

8

.

L.

Interference function 3

A polycrystalline thin film may be composed from a set of rectangular shaped crystallites of edges a, b and c. The variation of each crystallite may be accounted for by a log-norm distribution. Calculate the intensity / interference function ℑ(Q) for a Bragg reflection hkl. Solution Assuming the c axis of each crystallite is perpendicular oriented to the film surface, the column height distribution is P (L(x, y)) = dc . In this formula the crystal structure is assumed to be cubic with the cell constant d (not a, to avoid confusion with the major axis a). Insertion into Eq. (3.29) yields ℑ′rect (Q) =

c 2 1 b a sin (d· d ·Q/2) dxdy ab 0 0 sin2 (d·Q/2)

=

sin2 (cQ/2) b a ab sin2 (d·Q/2) 0 0

dxdy =

sin2 (cQ/2) sin2 (d·Q/2)

ab = 0b 0a dxdy (the area of a rectangle with the edges a and b) When each axis has a log-normal distribution interference function is: sin2 (cQ/2) grect (a, b, c) sin ℑrect (Q) = 2 (d·Q/2) dadbdc with the log-normal distribution of the crystallite shapes 3

grect (a, b, c) =

(2π)− 2 abc ln γa ln γb ln γc

2

For a cubic crystal structure Q may be

M.

(ln b−ln b0 )2 c−ln c0 )2 − (ln2(ln ) 2(ln γb )2 γc )2 √ 2π h2 +k2 +l2 . replaced by d

a−ln a0 ) · exp(− (ln2(ln − γa )2

.

Log-normal distribution function

Demonstrate the convergence of the log-norm distribution g(D) towards the delta function δ(D − D0 ) in the limit of ln γ → 0. Solution The delta function δ(D − D0 ) is (i) 0 for D = D0 and (ii) ∞ for D = D0 . (iii) The area under the delta function is 1. The latter property is also confirmed for g(D) by D0 )2 1 limD1 →0 D∞1 √2πD exp(− (ln D−ln )dD = 1 for ln γ > 0. 2(ln γ)2 ln γ The exponent

(ln D−ln D0 )2 2(ln γ)2

therefore (ii) g(D0 ) =

is zero for D = D0 , even in the limit of ln γ → 0,

√ 1 2πD0 ln γ

and limln γ→0 g(D0 ) = ∞;

(i) for all other positive D it becomes limln γ→0 g(D) = 0.

N.

Crude sample

Assume you dispose of a (rather abnormal) sample comprising of 2 × 109 cubes with one half having Dcub = 5 [nm] and the other half 10 [nm], when the cubic unit cell edges measures 0.5 [nm]. Give expression for p(L), g(D) and calculate Ln  and

Dn  for n = 1, 2, 3, 4. Solution 9

When p(L) is the probability for an arbitrarily picked column to have the size L and g(D) is the probability for an arbitrarily picked cube to have the size D the probablity for a certain L to be selected becomes p(L) = 15 δ(L − 5 nm) + 45 δ(L − 10 nm) because the larger cubes contain 20 × 20 columns and the smaller ones 10 × 10 columns. The crystallite size distribution function then is g(D) = 12 [δ(D − 5 nm) + δ(D − 10 nm)] because the numbers of large and small cubes are equal. Since f(x)δ(x − x0 )dx = f (x0 ) it is obtained

L1  = (5 [nm] +4 · 10 [nm])/5 = 9 [nm],

L2  = (25 [nm2 ]+4 · 100 [nm2 ])/5 = 85 [nm2 ],

L3  = (125 [nm3 ]+4 · 1000 [nm3 ])/5 = 825 [nm3 ],

L4  = (625 nm4 + 4 · 10000 nm4 )/5 = 8125 [nm4 ],

D1  = (5 [nm] +10 [nm])/2 = 7.5 [nm],

D2  = (25 [nm2 ]+100 [nm2 ])/2 = 62. 5 [nm2 ],

D3  = (125 [nm3 ] +1000 [nm3 ])/2 = 562. 5 [nm3 ],

D4  = (625 [nm4 ]+10000 [nm4 ])/2 = 5312. 5 [nm4 ].

O.

Integral intensities by weighting

In the early days of LPA x-ray diffraction patterns were obtained as strip charts from an x-t′ -plotter rather than as (2θi , Ii ) digital data sets. Integral intensities were obtained by sizing and weighting the Bragg reflections, while 2w and I0 parameters were derived from length measurements. Give some reasoning on the precision of Iint and β obtained by this procedure. Solution The Bragg reflections were to cut off from the plotter paper of the measurement and weighted in order to determine their mass Mhkl , which would scale with the integral intensity Ihkl . Evidently, the relative error ∆(Mhkl )/Mhkl would be the easier to minize the larger the peak was. Small peaks would thus be associated with a higher relative error than large ones. Therefore, it was recommended to make use of the full intensity scale when recording a peak. For alone standing peaks with not too large 2w this might have led to a relative error ∆(Ihkl )/Ihkl in the 1-5% range, which is not too different from the result of a numerical fit by virtue of a computer. The main problem in the achievable precision would occur for strongly overlapping peaks. In these cases, relative errors would easily amount to 10% and more, making the computer-based procedures definitely the favorite. It appears undecided, whether the computer-based determination of Ihkl is more time saving than the old swissor-based one. This holds in particular for users that only rarely make use of integral intensities and who have to care for an appropriate

10

program, installing it and learn to using it. It is evident, however, that the subsequent data processing is significantly simplified when data are already available in a computer-compatible format.

P.

Errata

The following mistypings were identified after publication to occur in the second chapter of the book (p. : page, L-x: line x from page bottom, Lx: line x from page top, Eq. : equation, Fig. : figure, sr: should read). p. 88, Eq. (i4.2): the square root has to be taken on the right hand side GoF = (...)1/2 p. 92, L-5: ”slightly steeper” sr ”slightly less steeper” p. 96, L-15: ”one the above” sr ”one of the above” p. 120, Eq. (3.49): is ”spheren” sr ”sphere” p. 137, L1: ”FWHU” sr ”FWHM”, L8: ”(3.25)” sr ”(3.26)”.

Q.

Acknowledgements

Many thanks is to Rainer Rudert for his assistance in solving the exercises and to Daniel Chateigner for identifying the errata.

11