- #1
TheOfficialAB
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Homework Statement
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Hi there!
This is a question from a practice problem sheet I got from the lecturer of my Condensed Matter 1 course.
Below are the normal vectors to the {111} and {112} lattice planes:
Homework Equations
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Bragg Condition: \begin{equation} n \lambda = d \sin (\theta) \end{equation}
Planar spacing, d: \begin{equation} d = \frac{a}{ \sqrt{ h^{2} + k^{2} + l^{2} } } \end{equation}
The Attempt at a Solution
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I have been able to calculate the angular positions of the θ peaks for both the cubic and tetragonal crystals.
a. First calculating the planar distances, (d), for both crystal types, and for both lattice planes.
For the cubic crystal, I took (a) as the unit cell length of the cube, 0.255 nm. Then I used it in the planar spacing equation with both normal vectors {111} and {112}. This resulted in the (d) spacings:
{111} Cubic
\begin{equation} \nonumber d = \frac{0.255 nm}{ \sqrt{ 1^{2} + 1^{2} + 1^{2} } } = \frac{0.255}{\sqrt{3}} = 0.147... \end{equation}
{112} Cubic
\begin{equation} \nonumber d = \frac{0.255 nm}{ \sqrt{ 1^{2} + 1^{2} + 2^{2} } } = \frac{0.255}{\sqrt{6}} = 0.104... \end{equation}
Then for the tetragonal crystal, I took (a) as the distance (c) given in the question, 0.257 nm. Then I used it in the planar spacing equation with both normal vectors {111} and {112}. This resulted in the (d) spacings:
{111} Tetragonal
\begin{equation} \nonumber d = \frac{0.257 nm}{ \sqrt{ 1^{2} + 1^{2} + 1^{2} } } = \frac{0.257}{\sqrt{3}} = 0.148... \end{equation}
{112}Tetragonal
\begin{equation} \nonumber d = \frac{0.257 nm}{ \sqrt{ 1^{2} + 1^{2} + 2^{2} } } = \frac{0.257}{\sqrt{6}} = 0.104... \end{equation}
b. I rearranged the Bragg Condition equation to give θ, with n = 1 (as I assume first order reflection).
\begin{equation} \theta = \arcsin{ \bigg( \frac{\lambda}{2 d} \bigg) } \end{equation}
and I used the four calculated values of (d) to find the four angular positions for each of the θ peaks.
{111} Cubic
\begin{equation} \nonumber \theta = \arcsin{ \bigg( \frac{0.1542 nm}{2 \times 0.147...} \bigg) } = 31.58^{\circ} \end{equation}
{112} Cubic
\begin{equation} \nonumber \theta = \arcsin{ \bigg( \frac{0.1542 nm}{2 \times 0.104 ...} \bigg) } = 47.78^{\circ} \end{equation}
{111} Tetragonal
\begin{equation} \nonumber \theta = \arcsin{ \bigg( \frac{0.1542 nm}{2 \times 0.148...} \bigg) } = 31.30^{\circ} \end{equation}
{112} Tetragonal
\begin{equation} \nonumber \theta = \arcsin{ \bigg( \frac{0.1542 nm}{2 \times 0.104...} \bigg) } = 47.29^{\circ} \end{equation}
c. It is at this point I find myself in difficulty. How does one tell if the tetragonal peaks are simply shifted cubic peaks, or if they have split into multiple peaks?
I have looked through my lecture notes, Ashcroft & Mermin, Kittel, and performed Google searches. I can't find much information on the splitting of x-ray diffraction peaks. This one link states that the {111} peak does not split after cubic to tetragonal phase transitions: http://pd.chem.ucl.ac.uk/pdnn/refine2/phase.htm
The difference in angular positions does not seem to be much greater than half a degree. Is this a large difference? I'm not familiar with the relative size of these changes.
I'm thinking this depends on the symmetry of the lattice plane in question, correct? Do lattice plane families also come into the picture here? If splitting occurs, does the peak split into the members of its family?
Any thoughts would be greatly appreciated.
Thank you for your time!