Exercises, Solutions and Errata to Chapter 1 of

A. v2

Exercises, Solutions and Errata to Chapter 1 of ”Thin Film Analysis by X-ray Scattering” Wiley-VCH, Weinheim, 2006 Mario Birkholz Thin Film Analysis and Technology Postfach 120610, 10596 Berlin, Germany

Abstract Chapter 1 introduces the basic phenomena of x-ray diffraction like angular position and intensity of Bragg peaks and the θ/2θ scan. The instrumental and structure boxes are related to the powder diffractometer, the generation of x-rays by sealed x-ray tubes and the basic one-elemental structures of the fcc, bcc and hcp type. The following exercises are related to these areas. Also an errata list is given for the first chapter.

Classical radius of the electron

The determination of the classical radius of the electron proceeds by assuming
that the total electric field energy E · DdV is confined to the space surrounding the spherical particle of radius re . Show that re = e2 /(4πε0 mc2 ) is obtained when the total field energy is identified with the energy of the rest mass mc2 . Solution. The energy stored in the field is derived from the volume integral over the product of the electrical field E and the dielectric displacement field D = ε0 E. The energy mc2 can thus be expressed by the integral  DEdV = mc2 The electron is preferably assumed to reside at the origin of a spherical coordinate system and the integration is performed by virtue of spherical coordinates dV = r2 sin θdφdθdr with the coordinate r extending from re to infinity. The integral then becomes
∞ (−e) (−e) 2 r dr 4π re ε0 4πε0 r2 4πε0 r2
∞ e2
e2 e2 ∞ = re r2 dr = re dr = = mc2 4πε0 r4 4πε0 r2 4πε0 re

B.

Optical length of coherence

Consider a preferably plane transparent thin film on a reflecting substrate (like a Si wafer). Examine it in daylight and in a room illuminated by neon light. Which optical phenomena can be observed in both cases? Why is the optical length of coherence longer in one case than in the other? How does the coherence length depend on the frequency spectrum of the radiation? Solution. If the transparent film exhibits a reliable thickness homogeneity colorful lines will be observed the color of which depending upon the optical path nℓ in the film with n standing for the refractive index. Since the effect relies on the effect of interference it will only be observed under neon light or other line emitters that comprise of a small spectral bandwidth ∆λ/λ. The effect will remain unobserved under daylight, since the solar spectrum is that of a broad band emitter. Interference phenomena always deserve a small line width for constructive and destructive interference to be observed. This holds for optical phenomena in thin films as for the diffraction of x-rays from a crystalline lattice.

C.

Momentum of an x-ray photon

What is the momentum of an 8 keV x-ray photon? Compare this to the average momentum of a Ar gas atom confined to a vessel under thermodynamic standard conditions. 1

2

Solution. According to deBroglie the momentum of a quantum mechanical particle is given by p = h/λ. This results in p = 6.626176 × 10−34 /(0.154 · 10−9 ) [Js/m] = 4. 30 × 10−24 [N/m] for an x-ray photon having a wavelength of λ = 0.154 [nm]. The momentum of an ideal gas atom is given by the equipartition theorem by 3 p = (2m kT )1/2 . Inserting m = 39.95u for Ar yields p = (2 · 39.95 · 1.660556 × 10−27 · 2 3 1.3807 × 10−23 · 293)1/2 [kg J]1/2 = 2.84 × 10−23 [N/m] for an Ar gas atoms under 2 standard conditions.

D.

Reciprocal lattice

Give the general expression for 1/d2hkl for the triclinic crystal (Table 1.1) for the triclinic crystal by starting from 1/d2hkl = (hb1 + kb2 + lb3 )2 , with the reciprocal lattice vectors bi depending like bi = (aj × ak )/(ai (aj × ak )) on the unit cell vectors ai . Solution. The magnitudes of the unit cell vectors a1 , a2 and a3 are identical with a, b and c. The angles between the unit cell vectors are defined by ∡(a, b) = γ, ∡(a, c) = β and ∡(b, c) = α. Inserting the three equations bi = (aj × ak )/(ai (aj × ak )) for each (i, j, k) combination, i.e. (1, 2, 3), (2, 1, 3) and (3, 1, 2) into 1/d2hkl = (hb1 + kb2 + lb3 )2 yields 1 = (h(a2 × a3 )/(a1 (a2 × a3 )) + k(a1 × a3 )/(a2 (a1 × a3 )) + l(a1 × a2 )/(a3 (a1 × a2 )))2 d2hkl The denominator in each term on the rhs ai (aj × ak ) is equal to the unit cell volume Vuc , for which the equality 1/2 Vuc = abc (1 + 2 cos α cos β cos γ − cos2 α − cos2 β − cos2 γ) holds in the general triclinc case. This leads to (h(a2 × a3 ) + k(a1 × a3 ) + l(a1 × a2 ))2 1 = 2 2 2 2 dhkl a b c (1 + 2 cos α cos β cos γ − cos2 α − cos2 β − cos2 γ) and   1 1 h2 (a2 × a3 )2 + k 2 (a1 × a3 )2 + l2 (a1 × a2 )2 + = 2 2 dhkl Vuc +2hk(a2 × a3 )(a1 × a3 ) + 2hl(a2 × a3 )(a1 × a2 ) + 2kl(a1 × a3 )(a1 × a2 ) The first three squares can be simplified by h2 (a2 ×a3 )2 +k 2 (a1 ×a3 )2 +l2 (a1 ×a2 )2 = h2 b2 c2 sin2 α+k 2 a2 c2 sin2 β +l2 a2 b2 sin2 γ The second three squares of the type 2hk(a2 × a3 )(a1 × a3 ) have to be solved by Lagrange’s identity (a × b)(c × d) = (ac)(bd) − (bc)(ad), which gives (a2 × a3 )(a1 × a3 ) = (a2 a1 )(a3 a3 ) − (a3 a1 )(a2 a3 ) = abc2 (cos α cos β − cos γ) (a2 × a3 )(a1 × a2 ) = (a2 a1 )(a3 a2 ) − (a3 a1 )(a2 a2 ) = ab2 c(cos α cos γ − cos β) (a1 × a3 )(a1 × a2 ) = (a1 a1 )(a3 a2 ) − (a3 a1 )(a1 a2 ) = a2 bc(cos β cos γ − cos α) One finally ends up with the expression for the inverse square of the lattice plane distance d−2 hkl in triclinic crystals 3



 k2 l2 hk h2 2 2 2  a2 sin α + b2 sin β + c2 sin α + 2 ab (cos α cos β − cos γ)+    hl kl +2 (cos γ cos α − cos β) + 2 (cos β cos γ − cos α) 1 ac bc = d2hkl 1 + 2 cos α cos β cos γ − cos2 α − cos2 β − cos2 γ From this expression the other relations in Table (1.1) may be derived by inserting the constraints for a, b, c, α, β and γ of the higher symmetry crystal systems. It should be noted that reciprocal lattice vectors bi are often defined via bi = 2π(aj × ak )/(ai (aj × ak )), which differs by the constant factor 2π from the expression given above. This ambivalence compares to the definition of the Fourier transformation and its reverse and illustrates the fact that the reciprocal space may be considered as the set of points that describe the Fourier transforms of all standing waves in the direct lattice. If the alternative definition of bi is used the starting equation of this exercise has to be modified obviously.

E.

Lattice factor of the simple cubic lattice

Show that the summation over all scattering centers according to Eq. (1.9) indeed yields Eq. (1.10). Make use of the geometrical sum. Solution. Eq. (1.10) is derived by multiplying (1.9) with its complex conjugate and evaluating the individual sums. For this purpose the threefold sum is preferably devised into indidual terms N 1 −1 N 2 −1 N 3 −1    exp (−iQ(n1 ac1 + n2 ac2 + n3 ac3 )) n1 =0 n2 =0 n3 =0 N 1 −1 

=

exp (−iQn1 ac)

n1 =0

N 2 −1 

exp (−iQn2 ac)

n2 =0

N 3 −1 

exp (−iQn3 ac)

n3 =0

Multiplying with the complex conjugate yields three factors of the form N−1

N−1   exp (−inx) exp (inx) n=0

n=0

where aQcj has been abbreviated by x. The summation yields according to the geometrical sum 1 − e−iNx − eiNx + 1 e−iNx − 1 eiNx − 1 · ix = e−ix − 1 e −1 1 − e−ix − eix + 1 cos Nx − 1 sin2 (Nx/2) 2 − (cos N x − i sin N x) − (cos Nx + i sin N x) = = = 2 − (cos x − i sin x) − (cos x + i sin x) cos x − 1 sin2 (x/2) Performing these step for each of the three terms in (1.9) gives the product of three Laue functions (1.10), i.e. x = aQcj = πh has to be obeyed for every direction cj for a reflection to be observed. A reflection maximum is thus observed for every triple of integer numbers hkl.

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F.

Atomic form factor

2

Derive the atomic form factor of He by modeling it with the spherical charge 3 2e 2 distribution of two 1s electrons ρe (r) = exp(−4r/a0 ) and performing the π a0 integration (1.17). Find a numerical approximation in the form of Eq. (1.18) through a non-linear regression. What is the precision of the fit in the sin θ/λ range from 0 to 10 nm−1 when the numerical value for Bohr’s radius a0 = 0.0529 nm is inserted? Solution. The volume integral over the spherical charge distribution gives the charge of the two He electrons:    
∞ 2e 2 3
2e 2 3 exp(−4r/a0 )dV = 4π 0 exp(−4r/a0 )r2 dr = 2e π a0 π a0 For the solution of the integral  f = ρe (r) exp(−iQr)dV the scattering vector Q is assumed to conincide with the positive z axis of the spherical charge distribution and the angle between Q and r is denoted by γ. The vector product Qr may then be expressed by Qr cos γ and it is thus derived for any spherical charge distribution
∞

∞
2 sin Qr 2 f = 2π 0 0 πρe (r) exp(−iQr cos γ) sin γr2 dγdr = 2π 0 0 πρe (r) r dr Qr Inserting the idealized He charge distribution results in  
∞ 2e 2 3 2 sin Qr 2 r dr exp(−4r/a0 ) fHe = 2π 0 π a0 Qr 32e 32e = = 2 a20 a40 a2 2 4 16 + 8Q +Q 4 + Q2 0 4 16 4 4π sin θ Making use of the definition of momentum transfer Q = gives λ −2  sin2 θ fHe = 2e 1 + π 2 a20 2 (1) λ Atomic form factors are usually given in units of elementary charges e. In the case outlined here, the function fHe /e has the appearance

1.5

1

0.5

00 2 2 4 6 8 10 x (1 + (0.0529πx)2 )2 For higher-Z atoms more complicated expressions arise and their form factors may not as simply be described as in the He case. Then, their form factors are approxi sin2 θ mated by specifying the a, b, c coefficients of the expansion f = c0 + 4j=1 aj exp(−bj 2 ) λ (given, for instance, in the International Tables for Crystallography, Vol. C). The coefficients of the fHe /e function given above can be found by a nonlinear regression and it is obtained c0 = 0.03051, a1 = 0.90158, b1 = 0.04733, a2 = 0.53509, b2 = 0.01649, a3 = 0.47594, b3 = 0.1006, a4 = 0.05688, b4 = 0.195. A plot of the coefficient function is practically indistinguishable from the plot given above. This example demonstrates the basic procedure by which atomic form factors f are calculated from the electron wave functions Ψ, since the square of the latter multiplied by the elementary charge e gives the electron density ρe . The different sets of atomic scattering factors to be found in the literature just differ by the degree of sophistication of the Hamiltonian, by which the ground state of the atom is modeled.

fHe /e =

G.

Structure factor from the lattice factor

Derive the structure factor Fh and extinction conditions of the three lattice types bcc, fcc, hcp from the general expression (1.23). Show that the evaluation of an expression comparable to Eq. (1.9) for the bcc lattice yields an equivalent result. Solution. Atoms in the bcc structure reside on lattice sites desribed by the lattice vector rmnp = mac1 + nac2 + pac3 3 5 1 having m, n, p = 0, ± , ±1, ± , ±2, ± , ... 2 2 2 Each sum over either m, n or p thus splits into a double sum N 1 −1 N 2 −1 N 3 −1    exp (−i(maQc1 + naQc2 + paQc3 )) m=0 n=0 p=0 N 1 −1 N 2 −1 N 3 −1   

=

exp (−ia(n1 Qc1 + n2 Qc2 + n3 Qc3 )) +

n1 =0 n2 =0 n3 =0

5

6

  1 1 1 exp −ia((n1 + )Qc1 + (n1 + )Qc2 + (n1 + )Qc3 ) 2 2 2 n1 =0 n2 =0 n3 =0    N 1 −1 N 2 −1 N 3 −1   i = 1 + exp − a(Qc1 + Qc2 + Qc3 ) exp (−ia(n1 Qc1 + n2 Qc2 + n3 Qc3 )) 2 n =0 n =0 n =0

+

N 1 −1 N 2 −1 N 3 −1   

1

2

3

The intensity distribution results from multiplying with the complex conjugate        3 sin2 (Nj xj /2) i i (x1 + x2 + x3 ) 1 + exp − (x1 + x2 + x3 ) 1 + exp 2 2 sin2 (xj /2) j=1

3   sin2 (Nj xj /2) = 2 1 + cos 12 (x1 + x2 + x3 ) sin2 (xj /2) j=1

The second factor allows for constructive interference if the Laue conditions as given in eq. (1.11) are obeyed, i.e. if x1 = aQc1 = πh etc. Therefore x1 + x2 + x3 = π(h + k + l). The first factor wil only yield a non-zero result for x1 +x2 +x3 = π(h+k+l) = 2πn, and for x1 + x2 + x3 = π(h + k + l) = 2π(n + 1) the reflection will vanish. These are the same conditions for constructive and destructive interference as given in eq. (1.25).

H.

Multiplicities

What are the multiplicities m(hkl) of Bragg peaks of the thin Al film shown in figure 1.10? Solution. m(111) = 8, m(200) = 6, m(220) = 12, m(311) = 24, m(222) = 8, m(400) = 6, m(331) = 24, m(420) = 24.

I.

alpha1 and alpha2

Suppose a Bragg reflection in a θ/2θ pattern to be observed at 2θ1 when Cu Kα1 radiation is applied. What would be the appropriate position of 2θ2 for the Cu Kα2 line? In absolute values for 2θ1 = 32.000◦ ? Solution. Both peaks arise at positions that are related to the same interplanar spacing d. This spacing is invariable and the solution to the problem derives from applying Bragg’s law for both peaks: d=

λ2 λ1 = 2 sin θ1 2 sin θ2

The index 1 or 2 stands for the α1 - or α2 -related peak and x-ray wavelength. This ansatz gives the relation   λ2 sin θ1 θ2 = arcsin λ1 7

between α1 and α2 peak positions. For usage of Cu Kα and a peak at 2θ1 = 32.000◦ the second peak would be observed at 2θ2 = 32.081◦ . The α1 − α2 -spacing is thus smaller than 0.1◦ in this scattering angle range.

J.

Positional shift of beta peaks

A diffractometer is run without suppression of Kβ radiation. Where do the Kβ peaks occur in the diffraction pattern compared to the α peaks? General: are the interference lines of higher energy radiation shifted towards lower or higher scattering angles 2θ? Solution. Kβ photons exhibit a larger energy (smaller wavelength) compared to Kα photons. According to same arguments as given in the previous exercise the Kβ lines are thus shifted to lower scattering angles compared to Kα lines. It may be a valuable exercise to scan a θ/2θ pattern of a thin film sample with and without Kβ filtering to get aware of the effects of the tube’s emission profile.

K.

c/a ratio in the hcp structure

Calculate the c/a ratios for the hcp-structured metals given in table s1.1. How do they deviate from the ”ideal” value of 1.633 and what is the meaning for the two different interatomic distances d1 and d2 in the hcp structure? Solution. c/a = 1. 624 for Mg, = 1. 654 for Ti, = 1. 8563 for Zn, and = 1. 593 for Zr. There occur two distinct bond distances d1 and d2 in the primitive hexagonal unit cell: one interatomic bond in the basal plane, having d1 = a, andanother bond dis c 2 . The tance d2 between atoms at (000) and ( 23 13 12 ), which becomes d2 = a 13 + 14  a   c 2 hcp structure derives from equating both bond lengths d1 = d2 ⇒ a = a 13 + 14 a  c yielding 8/3 = 1.633. For most one-elemental metals crystallizing in the = a hexagonal structure the unit cell deviates from the hexagonal closed-packed one exc hibiting a non-ideal = 1.633. However, the naming become widely used for all of a them.

L.

Nearest neighbors

Determine the distance d of an atom in the fcc and bcc structure to its nearest neighbors as a function of the lattice parameter a. Do the same for the hcp structure for interatomic distances within the basal plane and in the out-of-plane direction. Assume that atoms may be modeled by spheres and calculate the volume ratio of space occupied by spherical atoms Vat and the complete unit cell volume Vuc for all three structure types. 8

Solution. The solution in the fcc structure is found by considering the distance between the atom at (000) and a second one in the basal plane at ( 12 12 0) yielding √ √ d = 2a/2 = a/ 2. Since there are four atoms in the fcc unit cell the atomic-over-unit-cell ratio becomes     √ 3 2 a 4 d 4 4 /a3 = 3 π π( √ )3 = π = 0.7405. Vat /Vuc = 4 · 3 2 a 3 2 2 6 √ In the bcc structure d = 3a/2 and     √ 3 3 4 d 2 4 √ a 3 Vat /Vuc = 2 · π π( 3 ) = π = 0.6802. /a3 = 3 3 2 a 3 2·2 8 Assuming a spherical extension of atoms the bcc structure is thus less dense packed than the fcc structure. hcp structure: volume ratio in the primitive hexagonal cell Atom-over-unit-cell 3  3  √ 4 d2 4 d1 + π 2/ 3a2 c π Vat /Vuc = 3 2 3 2 3   8 π  a 3 8 π a 1 1  c 2 = √ + + √ 3 3a2 c 2 3 3a2 c 2 3 4 a

  3/2 1 1 c2 π a + 1+ = √ 2 3 4a 3 3c c  inserting the constraint for hcp = 8/3 = 1. 633 yields the numerical result a   3/2 √ π 3 2 1 18 Vat /Vuc = √ = + π = 0.7405 1+ 3 43 6 3 3 8 The hcp structure thus gives the same volume filling ratio of the unit cell as the fcc structure. As outlined in the structure box they are both close-packed structures and just differ by the way of stacking the atomic planes.

M.

Linear attenuation coefficient

Calculate the linear attenuation coefficient µ and the penetration depth τ1/e for Be, Zn, Fe, Ag and Au for irradiation with Cu Kα (8.04 keV). Determine the density and mass attenuation coefficient of the materials from tables like Table 2.1 or from the internet. Solution. The penetration is defined as the inverse attenuation coefficient τ1/e = µ−1 = (ρµm )−1 cm3 kg [10−3 m] = 4. 8697 × 10−3 [m] = Be: (ρµm )−1 = (1.85 · 0.111)−1 [ ]= g m2 1.85 · 0.111 4. 87 [mm]

9

cm3 kg [10−3 m] ]= = 24.2 [µm] g m2 7.14 · 5.79 3 −3 cm kg [10 m] ]= = 4. 21 [µm] Fe: (ρµm )−1 = (7.86 · 30.2)−1 [ g m2 7.86 · 30.2 3 −3 cm kg [10 m] ]= Ag: (ρµm )−1 = (x · y)−1 [ = x [µm] g m2 x·y 3 −3 cm kg [10 m] = x [µm] ]= Au: (ρµm )−1 = (x · y)−1 [ g m2 x·y Zn: (ρµm )−1 = (7.14 · 5.79)−1 [

N.

Beta filtering

The mass absorption coefficients for Cu Kα and Cu Kβ1 radiation in Ni are 48.8 and 279 cm2 /g while the mass density of Ni is 8.9 cm3 /g. Calculate the intensity ratios I(Kα)/I(Kβ1 ) and I(Kα)/I0 (Kα) after Cu K radiation with an initial ratio I0 (Kα)/I0 (Kβ1 ) of 3:1 has been transmitted through a Ni foil of (a) 8, (b) 15 and (c) 20 µm thickness. Which of these foils would you consider as the optimum Kβ filter? Solution. The x-ray beam passes the Ni foil under normal incidence. Therefore the intensity ratios are simply calculated by applying the law of Lambert-Beer I(Kα)/I(Kβ1 ) = I0 (α) exp(−µαt)/ (I0 (β) exp(−µβt)) = 3 exp(−µαt)/ exp(−µβt) (a) = 3 exp(−48.8 · 8.9 · 0.0008)/ exp(−279 · 8.9 · 0.0008) = 15. 45 (b) = 3 exp(−48.8 · 8.9 · 0.0015)/ exp(−279 · 8.9 · 0.0015) = 64. 83 (c) = 3 exp(−48.8 · 8.9 · 0.0020)/ exp(−279 · 8.9 · 0.0020) = 180. 6 I(α)/I0 (α) = exp(−µαt) (a) = exp(−48.8 · 8.9 · 0.0008) = 0.7065 (b) = exp(−48.8 · 8.9 · 0.0015) = 0.5213 (c) = exp(−48.8 · 8.9 · 0.0020) = 0.4195 The optimum filter thickness will depend on the accepted signal-to-noise or signal-to-background ratio in the measurement under consideration. For many measurements like those for chemical phase identification a suppression of the Kβ-line relative to the Kα one by a factor of 1:65 compared to the initial value of 1:3 is fully sufficient (case (b)), although the choice of 15 µm Ni foil is associated with a reduction of the irradiating Kα intensity by 48% or almost 50%. In consequence, the usage of the 15 µm Ni filter requires to extend the integration time per ∆(2θ) step by a factor of about 4 to end up with the same signal-to-noise ratio of the Kα Bragg peaks compared to a scan without Ni foil.

O.

Coincident reflections

In cubic substances two group of lattice planes √ may scatter simultaneously for the same scattering angle 2θ, because dhkl = h2 + k 2 + l2 /a is not a one-to-one unique function of Miller indices hkl. Which are the first double reflections to occur in a θ/2θ scan? Give the first three of them for fcc and bcc. 10

Solution. face-centered cubic: 333/511, 442/600, 711/551, body-centered cubic: 411/330, 510/431, 433/530.

P.

Number of photons

An x-ray copper tube is operated with 40 kV and 40 mA and is assumed to have an efficiency for x-ray generation of 1%. X-rays are emitted with cos2 characteristic and the take-off angle is 6◦ . The divergence slit is 1◦ . By how many x-ray photons is the sample irradiated per second? By how many Cu Kα photons? Give order of magnitude estimations. Solution. Running the x-ray tube with 40 [kV] and 40 [mA] is adequate to 1600 [W]. Converting this power completely into CuKα x-rays one would end up with 1600 1600 [s−1 ]= 1.2 × 1018 [s−1 ] [W/keV]= 8.04 8.04 · 103 · 1.6022 · 10−19 CuKα photons. However, due to physical reasons and geometrical dilution of radition this number becomes significantly reduced. Firstly, it may assumed that only 1% of the electrical power is converted into radiation, while the major part becomes thermalized and is taken away by the water cooling of the anode. This results in a reduction factor of 10−2 . Next, the spectral weight of the Kα radiation can roughly be estimated from the inspection of figure i2.2, figure 3.4 and table 3.2 to result in a further reduction factor of 10−3 . Considering the x-rays on their way from the anode focus to the sample, one realizes that one half penetrates into the anode, while only the other half enters the half sphere above the anode surface - yielding a reduction factor of 0.5. Only a part of the x-rays emitted from the anode focus into the half sphere above are extracted to irradiate the sample. The geometry is accounted for by spherical coordinates by azimuth φ and polar angle ϑ. The spherical surface area to be considered is restricted to φ0 + ∆φ and ϑ0 + ∆ϑ. Therefore, only the share
φ0 +∆φ
ϑ0 +∆ϑ sin ϑdϑdφ φ0 ϑ0 is extracted via the anode window. The extension ∆φ of the azimuth can be estimated from the width of the anode focus and a typical diffracometer radius amounting 180 10 to 10 and 300 mm, respectively, yielding ∆φ = arctan ≈ 2◦ π 300 ◦ Since the take-off angle and divergence slits are set to 6 and 1◦ , respectively, minimum and maximum ϑ values amount to 83.5◦ and 84.5◦ . It thus has to be calculated π
π 1
+1 180 84.5 180 sin ϑdϑdφ = 9. 64 × 10−5 ≈ 10−4 π π −1 180 83.5 180 2π where the prefactor stands for the normalization with respect to the total surface of a half sphere. This reduction factor accounts for the shieding and the dilution of the radiation from the anode to the sample. 11

One finally ends up with a total reduction factor of 0.5·10−2 ·10−3 ·10−4 ≈ 5·10−10 . This would predict only some 108 CuKa photons to reach the sample surface. An improved quantitative calculation yields a results of about 107 CuKα x-ray photons for the specified conditions.

Q.

Diffractometer vs. spectrometer

What is the varied measurement quantity in all kinds of spectrometers and why is it inadequate to denote a diffractometer as a spectrometer? The same arguments are valid in distinguishing a diffraction pattern from a spectrum. Solution. In a spectrometer the energy of particles or photons received by the detector is varied. It thus is an energy-dispersive device. In contrast, in a diffractometer the energy of the radiation remains constant and only the pattern of the scattered radiation in space is monitored. It may thus be named an angle-dispersive measurement set-up, but the term spectrum is inadequate to account for a diffraction pattern.

R.

Radius of the focussing circle

Derive the dependence of the radius of the focussing circle RF C on the diffraction angle θ and goniometer radius R. Solution. It is realized by inspection from figure i1.1 that the center of the focussing circle is always found to lie on the z axis of the goniometer. During a θ/2θ scan the center approaches ever closer towards the sample with the limiting values: θ = 2θ = 0 → RF C = ∞ and θ = 2θ = 90◦ → RF C = R/2. An isosceles triangle can be inscribed into the figure by drawing the same length RF C from the center of the focussing circle (i) to the sample surface (in direction of the surface normal) and (ii) to the position of the receiving slit. This triangle shows the validity of π RF C = RF C cos 2θ + R cos( − θ) = RF C cos 2θ + R sin θ 2 and RF C (1 − cos 2θ) = R sin θ yielding the desired relation RF C = R

sin θ 1 − cos 2θ

and the limiting values: RF C (0) → ∞ and RF C (π/2) = R/2.

S.

Polonium

The element Polonium has only found few technical applications, since its isotopes are all unstable with the most long-living ones decaying under emission of α particles. The most relevant application of Po is as initiator in nuclear fission bombs. Devices based on uranium, for instance, operate by the gun principle and 12

comprise two undercritical U masses that are shot into each other for ignition. The chain reaction of fission processes is reported to be accelerated by covering one of the masses with Po and the other by Be. The α particles emitted from Po cause the liberation of neutrons from Be nuclei that lead to an acceleration of the U fission reaction. About 2 × 104 nuclear weapons with the explosive power of more than 106 Hiroshima bombs are still stored in the arsenals of the atomic powers although they pledged in the Non-Proliferation Treaty of 1970 ”to pursue negotiations in good faith on effective measures relating to cessation of the nuclear arms race at an early date and to nuclear disarmament, and on a Treaty on general and complete disarmament under strict and effective international control” (Article VI). Various organizations exists, in which scientists are acting for disarmament and the abolition of nuclear weapons like the International Network of Scientists and Engineers for Global Responsibility (www.inesglobal.com), the Pugwash movement (pugwash.org), the Union of Concerned Scientists (ucsusa.org) and others. Visit the internet sites of these organizations and evaluate what scientists can do to assure the humane and useful application of their work.

T.

Errata

The following mistypings were identified after publication to occur in the first chapter of the book (p.: page, L-x: line x from page bottom, Lx: line x from page top, Eq.: equation, F: figure, sr: should read). p. 2, E1.1: K, K0  stands for the scalar product between K and K0 p. 3, L-1: ”Eq. (1.3)” sr ”Eq. (1.5)” p. 6, L11 & L19: ”vector product” sr ”scalar product” p. 9, first Eq. ”(1.15)” sr ”(1.14)” p. 10, L14 & L15: ”Eq. (1.14)” sr ”Eq. (1.15)” p. 18, Eq. (1.18): ”cj ” sr ”c” p. 19, L18: ”fi ” sr ”fn ” p. 19, L23: ”xn ac1 ” sr ”xn ac1 ” p. 20, L7: ”theses” sr ”these” p. 26, L18: ”to2θmax ” sr ”to 2θmax ” p. 28, L16: ”easy” sr ”easily” p. 29, L8: ”multiplicity m” sr ”multiplicity mh ” p. 29, Eq. (1.34): ”N1 a” sr ”Ni a” p. 35, L5: ”3-5” sr ”10”

U.

Acknowledgment Thanks to Daniel Chateigner for pointing me to most of these mistakes.

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