Chapter 2 Elastic Scattering of Waves

Chapter 2 Elastic Scattering of Waves

1. Experimental setting of X-ray diffraction

· Incident X-ray is a plane wave

(, t) = exp i(- wt), where is the propagation vector. ê ê= 2p/l

· X-ray interacts with electrons.

· The interaction is regarded as elastic scattering of photons, so that ê ê= ê ê= 2p/l. is the wave vector of the diffracted beam.

· // , //

· Define Scattering Angle = 2q

· D = -

êDê= 2s sinq
= 2 sin q 2p/l

= 4p sin q/l

2. Scattered wave from a small element of electron cloud

· At t1, the X-ray reaches a scattering center at , such that the E-field is (, t1) = exp i(- w t1)

· The X-ray is scattered from as a spherical wave, such that the E-field reaching the detector at t1+t is :

· xscatt(, t) = exp i [(-) - wt]
= exp i [(-) - wt].

· For a small sample, the variation of is small when changes. in the amplitude is regarded as unchanged.

· The variation of the phase (-) µ Dr'/l » Dr'/0.1 nm and changes very fast when Dr' changes. Hence the factor - in the phase cannot be regarded as a constant.

· xscatt(, t) = A' exp (i) exp i [(-) - wt]
= A' exp i(- wt) exp i[(-)]
= A' exp i(- wt) exp -i(D)

3. Scattering by one atom :
xscatt,1 atom(,t)

= A' exp i(-wt)exp(-iD) n()d,

where n() is the electron density.

n()d is the number of electrons inside the volume d.

4. Atomic form factor

· Let the center of the jth atom to be .

· Decompose as + , where is the position of the electron cloud element w.r.t. the position of the atom.

· xscatt,1 atom= A' exp i(-wt)´
exp-iD(+) nj()d

=A' exp i(-wt) exp (-iD)´

exp(-iD) nj()d,

Define fj =exp(-iD) nj()d as the Atomic Form Factor of the jth atom.

· xscatt,1 atom= A' fj exp i(-wt) exp (-iD)

· fj is determined by the electron density and D.

5. Diffraction by all atoms

· xscatt(,t) = A' exp i(-wt)

´fm exp-iD(n1+n2+n3+),

where n1, n2 and n3 scan over all integers (lattice points), and m scans over all atoms in the basis.

xscatt(,t) = A' exp i(-wt) [fm exp-iD]

´[exp(-in1D)][exp(-in2D)][exp(-in3D)]

· Define Structure Factor F = fm exp-iD, which is related to the structure of the basis.

The intensity of the scattering wave is:
I µ xscatt xscatt* = çxscatt2ç
= çA2ç çF2ç [exp(-in1D)][exp(-in1D)]*

´ [exp(-in2D)][exp(-in2D)]*

´[exp(-in3D)][exp(-in3D)]*

6. Bragg Condition

· Consider the sum exp(-inx) = e-ix

·
= =

= = =

· has maximum value when x/2 = mp, where m = 0, 1, 2,.. (integer).

· We may replace x by D, so that

I =çA2ççF2ç

· I is maximum when
D/2 = pnh,

D/2 = pnk,

D/2 = pnl,
where n, h, k and l are integers.

· Consider D = nh+nk+nl = n, where is a reciprocal lattice vector,

D/2 = n(h+k+l) /2 = pnh, and

D/2 = pnk,
D/2 = pnl,
Laue condition : D in this form gives maximum intensity of the scattered wave.

· From Laue condition : çDç= nçç.
Sec. B.1 Þ çDç= 4p sin q/l

Sec. A.9 Þ çç = 2p/d

4p sin q/l = 2pn/d, gives

When n = 1, h, k and l do not have common factor, 2d sinq = l

(= D) is ^ to the (hkl) planes, i.e. (hkl) planes are involved in the diffraction.

· For a set of lattice planes with a fixed spacing d and fixed l, n = 1 corresponds to the lowest q. n > 1 corresponds to larger q, namely, higher order diffraction.

· 2(d/n) sin q = l describes the diffraction from planes with separation smaller than d. These planes would contain lattice points only when a non-primitive lattice is used for discussion.

7. Real Lattice Space and Reciprocal Lattice Space

Reciprocal space / Real space
Fundamental lattice vectors / = 2p/t,
= 2p/t,
= 2p/t. / =2p/Vc,
=2p/Vc,
=2p/Vc.
Lattice vector / =h + k + l / = n1+ n2 + n3

Note: Vc = volume of a primitive unit cell of the reciprocal lattice = . The unit is [L]-3, the reciprocal of that of .

Exercise : The fundamental lattice vectors of B.C.C. : =a(+-)/2, =a(-++)/2, and

= a(-+)/2.

Show that the fundamental reciprocal lattice vectors are :

=2p(+)/a, =2p(+)/a, and

=2p(+)/a.

Exercise : Fundamental lattice vectors of s.c. lattice are: =a, =a, and = a. Show that the fundamental reciprocal lattice vectors are:

=2p/a, =2p/a, and =2p/a.

Exercise : Find the reciprocal lattice vectors of :

(a) A linear atom chain with atomic separation a.

(b) F.C.C. with =a(+)/2, =a(+)/2 and =a(+)/2.

8. Structure factor effect:

· I µ çx2çµ çFç2 […]2 […]2 […]2

· Laue condition ensures […]2 are maximized

· çFç2 also affects I

Example : Find F of FCC

Assuming that the condition of = is satisfied.

F = fm exp (-i)

= fm exp (-i)

=h + k + l

= um + vm + wm, where um, vm and wm £ 1

= 2p (h um + k um + l wm)

Then F = fm exp -i2p (h um + k um + l wm)

=f {1+exp-ip(h+k) + exp-ip(k+l)+ exp-ip(h+l)}

[Consider the four atoms at 000, 0, 0, 0. fm = f.]

(i) h,k,l all even, F = 4f Þ maximum

(ii) h,k,l all odd, F = 4f Þ maximum

(iii) 2 even, 1 odd, F =f (1+1-1-1) = 0

(iv) 2 odd, 1 even, F = 0.

Note : contributions of the (100)

and (200) planes cancel each other.

Example : Structure factor of CsCl.

A basis contains atoms at:

= 0, = (++)/2.

F = fm exp -i2p (h um + k um + l wm)

= fCs exp (0) + fCl exp -i2p (h/2 + k/2 + l/2)

= fCs + fCl exp -ip (h + k + l)

(i) h + k + l is odd, F = fCs - fCl Þ weak line (not completely cancelled)

(ii) h + k + l is even, F = fCs + fCl Þ strong line

For s.c. cubic, (100) planes give 2d sin q = l
Þ constructive interference / For CsCl, contribution from the (100) and (200) planes partially cancelled with each others Þ weakening of I

Exercise : For B.C.C., show that

F = f [1+exp-ip(h+k+l)]

9. Table of diffraction lines of cubic systems

Cubic
h2+k2+l2 hkl
S.C. F.C.C. B.C.C. Diamond

1 100 ------

2 110 --- 110 ---

3 111 111 --- 111

4 200 200 200 ---

5 210 ------

6 211 --- 211 ---

8 220 220 220 220

9 300,221 ------

10 310 --- 310 ---

11 311 311 --- 311

12 222 222 222 ---

13 320 ------

14 321 --- 321 ---

16 400 400 400 400

10. Ewald Construction

Step 1 : Draw the reciprocal lattice space

Step 2 : Draw the incident wave vector , with the end of the vector located at one reciprocal lattice point

Step 3 : Draw a circle with the other end of the vector as the center and a radius of ||.

Step 4 : If a reciprocal lattice points is cut by the circle, the vector connecting the point and the center of the circle would be the wavevector of the constructive scattered X-ray beam. The reciprocal lattice vector =h+k+l connecting the two reciprocal points as shown is perpendicular to the (hkl) planes given the strong diffraction peak.

11. Experimental techniques

· X-ray tube : hot e- bombard a metal target

· Bremstrahlung radiation : When electrons are decelerated, a continuous spectrum of X-ray radiation is generated.

· If the electrons knock out some electrons from the inner electron shells of the atoms, the electrons of the outer electrons shells would drop to fill up the vacancies. This gives Ka and Kb lines of radiation.

Example

Radiation / Cu Ka1 / Cu Ka2 / Cu Kb
Wavelength (nm) / 0.1540562 / 0.154439 / 0.1392218
Intensity / very strong / strong / weak

Structure of an X-ray tube

l Monochromators

Absorbing filters

Target / Filter / Transmittance
Mo / Zr / 0.29
Cu / Ni / 0.42
Co / Fe / 0.45
Fe / Mn / 0.48
Cr / V / 0.49

Single crystal filters

· Collimator

· Detectors : films, proportional counter, Geiger-Muller counter, scintillation counter, semiconductor detector.

Laue method :

· Single crystal sample

· White light (Bremstrahlung radiation)

· The shortest and longest wavelengths lmin and lmax give ||= 2p/lmin and ||= 2p/lmax

· The Ewald construction shows that all the reciprocal lattice points between the two spheres give strong peaks.

Powder method

· Powder sample (random orientation) and single l (monochromatic)

· A film is rolled in a cylindrical shape with the sample placed at the center

· The diffraction pattern consists of concentric rings. 2q =

Normal film

X-ray diffractometer

· q-2q scan : when the sample is moved by Dq, the detector is moved by 2Dq