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ucph>Tommy: Created page with "We will here investigate the effect of the nuclear structure factor in the Diffractio..."

2019-07-14T21:32:51Z

Created page with "We will here investigate the effect of the nuclear structure factor in the Diffractio..."

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We will here investigate the effect of the [[Diffraction from crystals#label-eq:structurefactor|nuclear structure factor]] in the [[Diffraction from crystals#label-eq:diffract|total diffraction cross section]] for Bravais lattices. However, as we shall see, some Bravais lattices can with advantage be described as a lattice with a basis.

=====Question 1=====
For a simple cubic lattice, atom distance $a$, the reciprocal lattice points are also simple cubic, with point distance $\tau_{(001)} = 2 \pi / a$. Calculate the squared structure factor $|F_{\rm N}({\boldsymbol\tau}_{hkl})|^2$ in units of the nuclear scattering length $b$. Which reflections are allowed (non-zero structure factors)? Do the allowed reflections have different structure factors?

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}

<figure id="fig:Sc_bcc_fcc"> [[File:Sc_bcc_fcc.png| thumb | 400px | <caption> Simple cubic (left), body centered cubic (centre), face centered cubic (right) lattices with their basisatoms. The reciprocal lattice vectors are $\mathbf{a}*=\frac{2\pi}{a} (1,0,0)$, $\mathbf{b}*=\frac{2\pi}{a} (0,1,0)$ and $\mathbf{c}*=\frac{2\pi}{a} (0,0,1)$ in all three cases.</caption>]] </figure>

The simple cubic structure only has basisvector ${\mathbf d}=(0,0,0)$ in the cartesian coordinate system defined in <xr id="fig:Sc_bcc_fcc">Figure %i</xr> by $\{ {\mathbf a},{\mathbf b},{\mathbf c} \}$. In general For Bragg scattering ${\mathbf q}={ \boldsymbol\tau}= h{\mathbf a}^* + k{\mathbf b}^* + l{\mathbf c}^* $. Hence e.g. ${ \boldsymbol\tau}_{(001)}=0\cdot{\mathbf a}^* + 0\cdot{\mathbf b}^* + 1\cdot{\mathbf c}^*={\mathbf c}^*$.

\begin{align}
|F^{sc}_N({\boldsymbol\tau}_{hkl})|^2
&= \left| \displaystyle\sum_{ {\mathbf d}}b_{\mathbf d} e^{i \boldsymbol\tau_{hkl}\cdot {\mathbf d}} \right|^2\label{dummyafksjhfkjash1}\\
&= \left| b e^{i \frac{2\pi}{a}(h\cdot 0 + k\cdot 0 + l\cdot 0) } \right|^2\label{dummyafksjhfkjash2}\\
&= |b|^2 \label{dummyafksjhfkjash3}
\end{align}

Since ${\mathbf q} \cdot {\mathbf d} = 0$ for any $q$ here, hence giving $|F_N|^2=|b|^2$ all reflections are allowed.
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=====Question 2=====
A body centered cubic (bcc) lattice can be described as a simple cubic lattice (side length $a$) with a second atom placed in the centre of the cube. The metallic elements Cr and Fe are two examples. Calculate the distance between nearest lattice sites.

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}

The bulk centered cubic structure has atoms at sites ${\mathbf d} \in \{ (0,0,0), a(\frac{1}{2},\frac{1}{2},\frac{1}{2})\}$ in the cartesian coordinate system defined in <xr id="fig:Sc_bcc_fcc">Figure %i</xr> by $\{ {\mathbf a},{\mathbf b},{\mathbf c} \}$. The nearest neighbor distance is

\begin{equation}\label{dummyafksjhfkjash}
d = \sqrt{\left( \dfrac{a\sqrt{2}}{2} \right)^2 + \left( \dfrac{a}{2} \right)^2 } = \dfrac{\sqrt{3}a}{2}
\end{equation}
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Calculate the expression for the nuclear structure factor of a bcc structure using $b = 1$.

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
The structure factor is

\begin{align}
|F_N^{bcc}({\mathbf \tau}_{(hkl)})|^2
&= \left| b \left( 1+ e^{i\frac{2\pi}{a}(h, k, l)\cdot (\frac{a}{2},\frac{a}{2},\frac{a}{2})} \right)\right|^2\label{dummyafksjhfkjas1h}\\
&= \left| b \left( 1+ e^{i\pi(h + k + l)}\right)\right|^2\label{dummyafksjhfkjashs}
\end{align}
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Which reflections $(h, k, l)$ are allowed? Do they have different structure factors?

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
Using $b=1$ we get $|F_N({\boldsymbol \tau}_{(hkl)})|^2 = |2b|^2 = 4$ for $h+k+l$ even and $|F_N({\boldsymbol \tau}_{(hkl)})|^2 = 0$ (disallowed) for $h+k+l$ odd. All allowed reflections have the same structure factor and constitute an $fcc$ lattice.
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The structure factors you have calculated using $b = 1$ are also called the geometrical structure factors. Can you think of a reason why?

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
The geometrical structure factor only has to do with the relative positions of the atoms.
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=====Question 3=====
A face centered cubic (fcc) lattice (side length $a$) has atoms in the cube corners and at the centre of the cube faces. One example of an fcc system is Al. Calculate the distance between nearest neighbour lattice sites.

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
The face centered cubic structure has atoms at sites ${\mathbf d} \in \{ (0,0,0), a(\frac{1}{2},\frac{1}{2},0),a(0,\frac{1}{2},\frac{1}{2}), a(\frac{1}{2},0,\frac{1}{2})\}$ in the cartesian coordinate system defined in <xr id="fig:Sc_bcc_fcc">Figure %i</xr> by $\{ {\mathbf a},{\mathbf b},{\mathbf c} \}$. The nearest neighbor distance is

\begin{equation}\label{dummy236186318}
d = \sqrt{ \left( \dfrac{a}{2} \right)^2 + \left( \dfrac{a}{2} \right)^2 } = \dfrac{a}{\sqrt{2}}
\end{equation}
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Calculate the expression for the geometrical structure factor of the fcc structure.

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
The structure factor is

\begin{align}\label{dummy31763912769}
|F_N^{fcc}({\mathbf \tau}_{(hkl)})|^2
&= \left| b \left( 1+ e^{i\frac{2\pi}{a}(h, k, l)\cdot (\frac{a}{2},\frac{a}{2},0)}+e^{i\frac{2\pi}{a}(h, k, l)\cdot (0,\frac{a}{2},\frac{a}{2})}+e^{i\frac{2\pi}{a}(h, k, l)\cdot (\frac{a}{2},0,\frac{a}{2})} \right)\right|^2\\
&= \left| b \left( 1 + e^{i\pi(h + k)} + e^{i\pi(k + l)} + e^{i\pi(h + l)}\right)\right|^2
\end{align}
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Which reflections are allowed? Do they have different structure factors?

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
The structure factor is $|F_N({\mathbf \tau}_{(hkl)})|^2 = |4b|^2$ for all $h,k,l$ even or odd. Else disallowed.The allowed positions constitute a $bcc$ lattice.
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=====Question 4=====
How many lattice sites (atoms) are present in the unit cell of the simple cubic, the bcc, and the fcc lattices, respectively?

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
In the simple cubic crystal there is $\frac{1}{8}\cdot8 = 1$ atom.<br>
In the bulk centered cubic crystal there are $\frac{1}{8}\cdot8 + 1 = 2$ atoms.<br>
In the face centered cubic crystal there are $\frac{1}{8}\cdot8 + \frac{1}{2}*6 = 4$ atoms.<br>
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=====Question 5=====
The hexagonal lattice is described by the lattice vectors ${\bf a} = a(1, 0,0)$, ${\bf b} = a(-1/2, \sqrt{3}/2, 0)$, and ${\bf c} = c(0, 0, 1)$ in orthogonal coordinates. Calculate the expression for the corresponding reciprocal lattice vectors.

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
<figure id="fig:Hcp"> [[File:Hcp.png| thumb | 400px | <caption> Hexagonal crystal structure. The black coordinate system is orthogonal, whereas the red axes are the hexagonal lattice vectors. The gray spheres correpond to the two atoms in the $hcp$ basis.</caption>]] </figure>

In orthogonal coordinates the hexagonal lattice vectors can be written

$ \begin{align}
{\mathbf a}_c &= a (1,0,0)\\
{\mathbf b}_c &= a \left( -\dfrac{1}{2},\dfrac{\sqrt{3}}{2},0 \right)\\
{\mathbf c}_c &= c(0,0,1)\\
\end{align}$

Since $V_0= {\mathbf a}\cdot {\mathbf b}\times{\mathbf c}= \frac{\sqrt{3}}{2}a^2c $, the reciprocal lattice vectors in the cartesian coordinate system are

$ \begin{align}
{\mathbf a}_c^* &= \dfrac{2\pi}{V_0}{\mathbf b}\times {\mathbf c} = \dfrac{2\pi}{a} \left( 1,\dfrac{1}{\sqrt{3}},0 \right)\\
{\mathbf b}_c^* &= \dfrac{2\pi}{V_0}{\mathbf c}\times {\mathbf a} = \dfrac{2\pi}{a} \left( 0,\dfrac{2}{\sqrt{3}},0 \right)\\
{\mathbf c}_c^* &= \dfrac{2\pi}{V_0}{\mathbf a}\times {\mathbf b} = \dfrac{2\pi}{c}(0,0,1)\\
\end{align}$

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=====Question 6=====
The hexagonal closed-packed (hcp) lattice can be described as a hexagonal unit cell with a two-atom basis, where ${\bf d}_1 = (0,0,0)$ and
${\bf d}_2 = (a/2, a/(2 \sqrt{3}), c/2)$ in cartesian coordinates. Calculate the ratio $c/a$, knowing that the distance between an atom and all 12 nearest neighbours must be equal.

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
The hcp structure can be described by a hexagonal unit cell with a two-atom basis. In cartesian coordinates the atom positions are

\begin{equation}
{\mathbf d}_1 = (0,0,0) \qquad {\mathbf d}_2 = \left( \dfrac{a}{2},\dfrac{a}{2\sqrt{3}},\dfrac{c}{2} \right)
\end{equation}

The nearest neighbor distance is $a$ for all atoms, hence

\begin{align}
a &= |{\mathbf d}_2 - {\mathbf d}_1| = \sqrt{\left( \dfrac{a}{2} \right)^2 + \left( \dfrac{a}{2\sqrt{3}} \right)^2 + \left( \dfrac{c}{2} \right)^2} \Leftrightarrow \\
\dfrac{c}{a} &= 2 \sqrt{\dfrac{2}{3}} \approx 1.633
\end{align}
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In the hexagonal crystal coordinates, the relative atom coordinates may be given as $(0,0,0)$ and $(2/3, 1/3, 1/2)$, ''i.e.'' in units of the three real-space lattice vectors. Explain how the value $(2/3, 1/3, 1/2)$ appears.

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
In hexagonal crystal coordinates the lattice vectors are

\begin{equation}
{\mathbf a}_h = a(1,0,0), \quad {\mathbf b}_h = a(0,1,0), \quad {\mathbf c}_h = c(0,0,1)
\end{equation}

Then the position of the basisatoms are

\begin{equation}
{\mathbf d}_1 = (0,0,0)_c = (0,0,0)_h \qquad
{\mathbf d}_2 = \left( \dfrac{a}{2},\dfrac{a}{2\sqrt{3}},\dfrac{c}{2} \right)_c = \dfrac{2}{3}{\mathbf a}_h + \dfrac{1}{3}{\mathbf b}_h +\dfrac{1}{2}{\mathbf c}_h = \left( \dfrac{2}{3},\dfrac{1}{3},\dfrac{1}{2} \right)_h
\end{equation}
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=====Question 7=====
Calculate the expression for the structure factor of the hcp lattice using $b = 1$ and the atom positions in the hexagonal cell. Use the expression to calculate the geometrical structure factors for $(hkl) = (100)$, $(101)$, $(102)$, $(103)$, $(001)$, $(002)$, $(110)$, $(111)$ and $(112)$. Explain the regularities in the structure factors.

{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
In the hexagonal crystal coordinates the basisatoms are at

\begin{align}
{\mathbf d}_1 &= (0,0,0)\\
{\mathbf d}_2 &= \left( \dfrac{2}{3},\dfrac{1}{3},\dfrac{1}{2} \right)= \dfrac{2}{3}{\mathbf a} + \dfrac{1}{3}{\mathbf b} + \dfrac{1}{2}{\mathbf c}
\end{align}

A general position in the realspace crystal is ${\mathbf d} = n_a{\mathbf a}+n_b{\mathbf b}+n_c{\mathbf c} $.
The reciprocal lattice vectors in hexagonal crystal coordinates are

\begin{align}
{\mathbf a}^* &= \dfrac{2\pi}{a}{\mathbf b}\times {\mathbf c} = \dfrac{2\pi}{a} (1,0,0)\\
{\mathbf b}^* &= \dfrac{2\pi}{a}{\mathbf c}\times {\mathbf a} = \dfrac{2\pi}{a} (0,1,0)\\
{\mathbf c}^* &= \dfrac{2\pi}{c}{\mathbf a}\times {\mathbf b} = \dfrac{2\pi}{c}(0,0,1)\\
\end{align}

The scattering vectors can be written ${\boldsymbol \tau}$ = $h {\mathbf a}^* + h {\mathbf b}^* + k {\mathbf c}^* \equiv (h k l)$, hence the structure factors are

\begin{equation}
\left|F(hkl)\right|^2 = \left| \displaystyle\sum_{ {\mathbf d}}b_{\mathbf d} e^{i {\boldsymbol \tau}_{\rm (hkl)}\cdot {\mathbf d}} \right|^2 = \left| \displaystyle\sum_{n_an_bn_c}b_{n} e^{i 2\pi (n_a h + n_bk + n_c l)} \right|^2 = \left| 1 + e^{i 2\pi (\frac{2}{3} h + \frac{1}{3}k + \frac{1}{2} l)} \right|^2
\end{equation}

And specifically

\begin{align}
\left|F(100)\right|^2 &= \left| 1 + e^{i (\frac{4\pi}{3})} \right|^2 = \left( 1+\cos\left(\dfrac{4\pi}{3}\right)+ i\sin\left(\dfrac{4\pi}{3}\right)\right)\left( 1+\cos\left(\dfrac{4\pi}{3}\right)- i\sin\left(\dfrac{4\pi}{3}\right)\right) = (1-0.5)^2 + 0.75 = 1\\
\left|F(101)\right|^2 &= \left| 1 + e^{i 2\pi(\frac{7}{6})} \right|^2 = \left( 1+\cos\left(\dfrac{7\pi}{3}\right)+ i\sin\left(\dfrac{7\pi}{3}\right)\right)\left( 1+\cos\left(\dfrac{7\pi}{3}\right)- i\sin\left(\dfrac{7\pi}{3}\right)\right) = (1+0.5)^2 + 0.75 = 3\\
\left|F(102)\right|^2 &= \left| 1 + e^{i 2\pi(\frac{5}{3})} \right|^2 = \left( 1+\cos\left(\dfrac{10\pi}{3}\right)+ i\sin\left(\dfrac{10\pi}{3}\right)\right)\left( 1+\cos\left(\dfrac{10\pi}{3}\right)- i\sin\left(\dfrac{10\pi}{3}\right)\right) = (1-0.5)^2 + 0.75 = 1\\
\left|F(103)\right|^2 &= \left| 1 + e^{i 2\pi(\frac{13}{6})} \right|^2 = \left( 1+\cos\left(\dfrac{13\pi}{3}\right)+ i\sin\left(\dfrac{13\pi}{3}\right)\right)\left( 1+\cos\left(\dfrac{13\pi}{3}\right)- i\sin\left(\dfrac{13\pi}{3}\right)\right) = (1+0.5)^2 + 0.75 = 3\\
\left|F(001)\right|^2 &= \left| 1 + e^{i \pi} \right|^2 = 0\\
\left|F(002)\right|^2 &= \left| 1 + e^{i 2\pi} \right|^2 = 4\\
\left|F(110)\right|^2 &= \left| 1 + e^{i 2\pi(\frac{2}{3}+\frac{1}{3})} \right|^2 = 4\\
\left|F(111)\right|^2 &= \left| 1 + e^{i 2\pi(\frac{2}{3}+\frac{1}{3}+\frac{1}{2})} \right|^2 = 0\\
\left|F(112)\right|^2 &= \left| 1 + e^{i 2\pi(\frac{2}{3}+\frac{1}{3}+1)} \right|^2 = 4\\
\end{align}

Hence for $h=k$: $F^2=0$ for $l$ odd and $F^2=4$ for $l$ even . For $k+k$ odd: $F^2=3$ for $l$ odd and $F^2=1$ for $l$ even.

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Problem: Bragg scattering from Bravais lattices - Revision history

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