Crystal phase transition, Diffraction peak splitting problem

Tags:
1. Nov 8, 2015

TheOfficialAB

1. The problem statement, all variables and given/known data

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:

2. Relevant equations

Bragg Condition: $$n \lambda = d \sin (\theta)$$

Planar spacing, d: $$d = \frac{a}{ \sqrt{ h^{2} + k^{2} + l^{2} } }$$

3. The attempt at a solution

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
$$\nonumber d = \frac{0.255 nm}{ \sqrt{ 1^{2} + 1^{2} + 1^{2} } } = \frac{0.255}{\sqrt{3}} = 0.147...$$

{112} Cubic
$$\nonumber d = \frac{0.255 nm}{ \sqrt{ 1^{2} + 1^{2} + 2^{2} } } = \frac{0.255}{\sqrt{6}} = 0.104...$$

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
$$\nonumber d = \frac{0.257 nm}{ \sqrt{ 1^{2} + 1^{2} + 1^{2} } } = \frac{0.257}{\sqrt{3}} = 0.148...$$

{112}Tetragonal

$$\nonumber d = \frac{0.257 nm}{ \sqrt{ 1^{2} + 1^{2} + 2^{2} } } = \frac{0.257}{\sqrt{6}} = 0.104...$$

b. I rearranged the Bragg Condition equation to give θ, with n = 1 (as I assume first order reflection).

$$\theta = \arcsin{ \bigg( \frac{\lambda}{2 d} \bigg) }$$

and I used the four calculated values of (d) to find the four angular positions for each of the θ peaks.

{111} Cubic
$$\nonumber \theta = \arcsin{ \bigg( \frac{0.1542 nm}{2 \times 0.147...} \bigg) } = 31.58^{\circ}$$

{112} Cubic
$$\nonumber \theta = \arcsin{ \bigg( \frac{0.1542 nm}{2 \times 0.104 ...} \bigg) } = 47.78^{\circ}$$

{111} Tetragonal
$$\nonumber \theta = \arcsin{ \bigg( \frac{0.1542 nm}{2 \times 0.148...} \bigg) } = 31.30^{\circ}$$

{112} Tetragonal
$$\nonumber \theta = \arcsin{ \bigg( \frac{0.1542 nm}{2 \times 0.104...} \bigg) } = 47.29^{\circ}$$

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.

2. Nov 10, 2015

Oxvillian

Hi TheOfficialAB!

Your planar spacing formula (2) only applies to a simple cubic lattice. It looks to me like you've used it for the tetragonal lattice too.

3. Nov 11, 2015

TheOfficialAB

You're right! I did do the wrong calculations.
The correct formula for the d-spacing in the tetragonal phase appears to be:

$$d = \frac{a}{ \sqrt{ h^{2} + k^{2} } } + \frac{c}{ \sqrt{ l^{2} } }$$

I can use the fact that the volume remains constant after the phase change to calculate the side length of the 'square face' in the tetragonal.

$$\nonumber \sqrt{ \frac{(0.255 nm)^{3}}{c}} = 0.254 nm$$

I can then recalculate the {111} and {112} planes in the tetragonal phase

Tetragonal {111}
$$\nonumber d = \frac{0.254}{ \sqrt{ 1^{2} + 1^{2} } } + \frac{0.257}{ \sqrt{ 1^{2} } } = 0.4366...$$

$$\nonumber \theta = \arcsin { \bigg( \frac{0.1542 nm}{ 2 \times 0.436...} \bigg) } = 10.17^{\circ}$$

Tetragonal {112}
$$\nonumber d = \frac{0.254}{ \sqrt{ 1^{2} + 1^{2} } } + \frac{0.257}{ \sqrt{ 2^{2} } } = 0.36133...$$

$$\nonumber \theta = \arcsin { \bigg( \frac{0.1542 nm}{ 2 \times 0.36133...} \bigg) } = 12.32^{\circ}$$

However, though I get a greater discrepancy between the cubic and tetragonal angular positions now,
I'm still not seeing this splitting situation particularly clearly. How can I tell splitting should occur for one of the peaks and not the other?

Before you pointed out the fault in my tetragonal d spacing, I was using the same formula to calculate two d-spacings. One with (a) and one with (c).
I thought the tetragonal phase could be looked at as two different sets of planes with two different spacings, thus giving rise to splitting.
However the actual d-spacing formula nullifies that line of thinking because both (a) and (c) are used to calculate one d-spacing that is consistent.

4. Nov 11, 2015

Oxvillian

Hmm I think your plane spacing formula (4) is still not correct. It should come from
$$d = \frac{2\pi}{|{\bf G}|}$$
where ${\bf G}$ is the reciprocal lattice vector in question.

5. Nov 12, 2015

TheOfficialAB

Though the plane spacing can be related to the reciprocal lattice vector along the plane as you have said (proof of this is also in my notes), I have been given the Miller indices of lattice planes in this question. If I recall correctly the Miller indices are the shortest reciprocal lattice vector to normal to a lattice plane, w.r.t. to a set of primitive reciprocal lattice vectors (or simply the reciprocals of the intersects of the plane with the axes) which is what you have defined as G, at least I think so. If I had the reciprocal lattice vector in question, I could use the above formula, but not with what I have currently been given.

The formula (4), I couldn't find in my textbooks, but was available in several locations online:
http://web.eng.fiu.edu/~chenj/fall2008EMA5507C/Lecture8_Lattice2SpaceGroupNov7.pdf

I may also have come across a final solution? (some verification on this would be nice, if possible)

Since the (111) plane is symmetric, no splitting will occur after phase transition. If one looks at equation (4) then it can be seen that the denominator will be constant, no matter what arrangement of the family of {111} is used.

However (112) will not be symmetric in the tetragonal phase. Since (a) is not equal to (c), then a member of the family {112}, the direction (211) will show splitting. In this case we can see that using the equation (4), the plane (211) will present a different d-spacing than that of (112), using the the logic of the 22 term switching denominators from being under (c) to being under (a).

Previously I calculated the angular position of the (112) plane:

$$\nonumber \theta = \arcsin { \bigg( \frac{0.1542 nm}{ 2 \times 0.36133...} \bigg) } = 12.32^{\circ}$$

The plane (211) will have a different d-spacing:

$$\nonumber d = \frac{0.254 nm}{ \sqrt{ 2^{2} + 1^{2} } } + \frac{0.257 nm}{ \sqrt{1^{2} } } = 0.37059...$$

$$\nonumber \theta = \arcsin{ \bigg( \frac{0.1542 nm}{2 \times 0.37059...} \bigg) } = 12.01^{\circ}$$

Thus, we have splitting in the {112} direction after a cubic to tetragonal phase transition. The positions of the satellite peaks are 12.32° and 12.01°.

I am still unsure of one thing here, and my question is: what are the members of the {112} family? Do the members of the family change after the cubic to tetragonal transition? I know that in cubic phase, if the same three indices are present in the direction, regardless of sign (+ or -) then they are members of the same family. I'm not sure if this applies across other phases due to their change in symmetry.

Last edited: Nov 12, 2015
6. Nov 12, 2015

Oxvillian

The formula in your reference is correct, but it's not the same as your formula (4), basically because

$$\frac{1}{x+y} \neq \frac{1}{x} + \frac{1}{y}.$$

Sorry to keep complaining about this, but you can't really do the rest of the question without having the right plane spacing formula on the table!

It's really not that hard to derive - the reciprocal lattice vector in question should be very easy to write down.

The usual terminology is as follows: a family of lattice planes is just the set of parallel lattice planes corresponding to a given (hkl). In a cubic lattice, (112) and (121) are said to be equivalent families, because you can rotate the crystal so as to take one family into the other. In a tetragonal lattice, those two families aren't equivalent any more.

7. Nov 30, 2015

TheOfficialAB

Apologies for this late reply (and my lack of understanding of basic maths). Thanks for all your time and effort Oxvillian.

The due date for this practice problem is now well passed. However, after your last reply I saw where I was going wrong and fixed the calculations to reflect the correct d-spacing formula for the tetragonal phase. My conclusions seemed to check out, and I was able to submit the corrected version to the professor.

To quiet the OCD demons in my head, and for the sake of completeness, I will lay out my final solution as I have understood it below.

If there are still any areas where you feel like I have not properly understood the underlying theory, I really would appreciate it if you could tell me where they are!
I will try to derive the d-spacing formula as you suggested earlier on.

The correct d-spacing for the tetragonal phase is given by:

$$d = \frac{ 1 }{ \sqrt{ \frac{ h^{2} + k^{2} }{ a^{2} } + \frac{ l^{2} }{ c^{2} } } }$$

This results in the following tetragonal d-spacings and angular positions:

Tetragonal (111)
$$\nonumber d = 0.147...$$
$$\nonumber \theta \approx 31.59 ^{\circ}$$

Tetragonal (112)
$$\nonumber d = 0.1045...$$
$$\nonumber \theta \approx 47.54 ^{\circ}$$

Tetragonal (121)
$$\nonumber d = 0.103...$$
$$\nonumber \theta \approx 47.90 ^{\circ}$$

Conclusion
The tetragonal (111) peak (θ = 31.59°) was seen to simply be a shifted cubic (111) peak (θ = 31.58°).
The cubic (112) peak (θ = 47.78°) splits into two tetragonal satellite peaks (112) and (121, equivalently 211) which give (θ = 47.54°) and (θ = 47.90°) respectively.

8. Nov 30, 2015

Oxvillian

No problem - fwiw the main thing I noticed about this problem is that the (111) shift is a second-order shift - probably too small to see in a real experiment.

In other words, if you deform the original cubic crystal by a small amount $\epsilon$ and expand $\delta \theta$ as a Taylor series in $\epsilon$, you should find that the linear term vanishes.