MIT 8.231 Problem 2.1 Americium

March 28, 2015

Problem

The figure below shows a primitive unit cell of one crystalline form of the element√ americium. The space lattice is hexagonal with ~a1 = a~x,~a2 = 1/2a~x + 1/2 3a~y, and ~a3 = c~z. The basis is (0 0 0) , (2/3 2/3 1/4), (0 0 1/2) and (1/3 1/3 3/4).

a) Find the reciprocal lattice vectors G~ . Describe in words and sketch the reciprocal lattice. b) Sketch the first Brillouin zone. Give values for the important dimensions. c) Find the structure factors associated with the points (100), (001), and (120) of the reciprocal lattice. d) What is the ratio of the scattering intensity corresponding to (120) to that corresponding to (100)?

1 Solution a) ~ ~ ~ At first we want to calculate reciprocal lattice vectors b1, b2, b3. The vectors are defined as followed:

~ a2 × a3 b1 = 2π (1) a1(a2 × a3) ~ a3 × a1 b2 = 2π (2) a1(a2 × a3) ~ a1 × a2 b3 = 2π (3) a1(a2 × a3)

With the given values of ~ai we get:

√  3 ~ 2π b1 = √ −1 (4) 3a 0 0 ~ 4π b2 = √ 1 (5) 3a 0 0 2π b~ = 0 (6) 3 c   1

The reciprocal lattice vector is defined as followed: ~ ~ ~ G~ = hb1 + kb2 + lb3 (7)

Where h,k,l are Miller indices, with those indices Crystal planes are defined. G~ is also used for x-ray diffraction, the condition for diffraction is defined as: ~ ~ ~ G~ hkl = ∆k = k − k0 (8)

This is best shown with the Ewald’s Sphere. So the reciprocal lattice vector is used for calculating the X-ray diffraction of a certain plane of the crystal. The intensity of the diffraction is proportional to the squared Fouriercoefficients of the electron density, the Coefficient are defined as:

1 X ~ n = f (G)e−iG· ~rj (9) G~ V j j

2 Figure 1: 2D Sketch of a hexag- onal reciprocal lattice

So the reciprocal lattice vector for certain planes is connected with the intensity of the X-Ray diffraction.

In Figure 1 a 2D reciprocal lattice is shown, the lattice in 3D would look like more layers of the 2D lattice with the distance c to each other. The vector ~ b3 is placed placed normal to the plane of projection.

3 Figure 2: 2D Sketch of a hexag- onal reciprocal lattice with Bril- louin Zone drawn as dotted line b) In figure 2 we can see the drawn Brillouin Zone in the middle, the length from 4π 2π the center to each corner is 3a the height of the 3D Brillouin Zone is c . The structure of the Brillouin Zone is also hexagonal.

4 c) For finding the structure Factors we need the electron density, because no den- sity is given we take an approximation and say that every atom in the base has a density of Zδ(~r), the atomic number of Americium is Z=95.

For the calculation we want to use equation 9. As we see there is multi- plication of G~ and ~r, which have different bases, we can easily show that the ~ multiplication between ~ai and bj gives:

aibj = 2πδij (10)

We use equation 9 to calculate the structure factor, the rj are the bases given. The calculation for the (100 )plane goes as followed: 1 X −iG~ 100· ~rj n ~ = Zδ(~r − ~rj)e (11) G100 V j −2πi0 −2πi 2 −2πi0 −2πi 1 = Ze + Ze 3 + Ze + Ze 3 (12) = 95(1 − 0.5 − 0.866i + 1 − 0.5 + 0.866i) (13) n = 95 (14) G~ 100 1 The pre factor is V = 1 For the two other planes the calculation goes the same way: 1 X −iG~ 001· ~rj n ~ = Zδ(~r − ~rj)e (15) G001 V j −2πi0 −i π −πi0 −i 3π = Ze + Ze 2 + Ze + Ze 4 (16) n = 0 (17) G~ 001

1 X −iG~ 120· ~rj n ~ = Zδ(~r − ~rj)e (18) G120 V j = Ze−2πi0 + Ze−4πi + Ze−2πi0 + Ze−2πi (19) n = 380 (20) G~ 120

The results have been compared to numerical calculated structure factors you get from http://lamp.tu-graz.ac.at/~hadley/ss1/crystaldiffraction/atomicformfactors/ structurefactor.php with the values a=3.468 A and c=1.124 A:

|n | = 83.54 (21) G~ 100 |n | = 3.070 · 10−7 (22) G~ 001 |n | = 289.8 (23) G~ 120

5 Figure 3: Approximated form factor used for the numeric cal- culation from: http://lamp. tu-graz.ac.at/~hadley/ ss1/crystaldiffraction/ atomicformfactors/ formfactors.php

As you can see for the (001) our result is the same as numerical result, but for the (100) and the (120) plane the results don’t correspond, this because of the approximation for the density we made. The used form factor for the numeric calculation can be seen in Figure 3. d) The scattering intensity is the square of the form factors, so the ratio is:

2 |n ~ | G120 = 16 (24) |n |2 G~ 100

Philip Zettl

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