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Crystallography

A1  0.30 Write the formula 1 using the scattering vector $q$ ($q \ll k_i$).

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$
I = \frac{I_0}{N^2} \cdot \left( \frac{ \sin \left( Nqa/2 \right)}{ \sin \left( qa/2 \right) } \right)^2 \cdot \left( \frac{ \sin \left( qb/2 \right)}{\left( qb/2 \right) } \right)^2
$
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A2  0.20 Find scattering vector $q$ for the maximum numbered $h$ for a diffraction grating with a period $a$.

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$q = \frac{2\pi}{a} \cdot h, \quad h \in \mathbb{Z}$

DISCLAIMER. Here and below we use the notations:
$\mathbb{Z}$ – integer numbers: $\{..., -2, -1, 0, 1, 2, ...\}$
$\mathbb{N}$ – positive numbers: $\{1, 2, ...\}$
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A3  0.20 Let $ q_1 $ be the scattering vector for the first maximum. Express $ q $ in terms of $ q_1 $ for intensity maxima. How are $ q_1 $ and $ a $ related?

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$q = q_1 \cdot h, \quad h \in \mathbb{Z}$ 0.10
$q_1 \cdot a = 2\pi$ 0.10
A4  1.00 Observe the diffraction of samples DG1-DG5. Determine experimentally $ q_1 $, and $ a $ for each sample. Draw a scheme of your setup, write the quantities you measure, and write down the formulas for the calculations.

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Scheme of setup 0.10
Formula for calculation $q_1 = \frac{2\pi s_1}{L \lambda}$ or $a=\frac{\lambda L}{s_1}$, $s_1$ is distance between the two nearest reflexes on screen. 0.10
DG1: $ q_1 = 315 \pm 22 \mathrm{ mm^{-1}} $ 0.08
DG1: $a = 19.7 \pm 1.3 \mathrm{ \mu m} $ 0.08
DG2: $ q_1 = 125 \pm 10 \mathrm{ mm^{-1}} $ 0.08
DG2: $ a = 49.2 \pm 3.2 \mathrm{ \mu m} $ 0.08
DG3: $ q_1 = 78.5 \pm 5.5 \mathrm{ mm^{-1}} $ 0.08
DG3: $a = 78.8 \pm 5.2 \mathrm{ \mu m} $ 0.08
DG4: $ q_1 = 78.5 \pm 5.5 \mathrm{ mm^{-1}} $ 0.08
DG4: $a = 78.8 \pm 5.2 \mathrm{ \mu m} $ 0.08
DG5: $ q_1 = 78.5 \pm 5.5 \mathrm{ mm^{-1}} $ 0.08
DG5: $a = 78.8 \pm 5.2 \mathrm{ \mu m} $ 0.08
A5  1.50 Conduct an experiment and determine the $ a / b $ ratio for samples DG3, DG4, DG5. Explain your solution using formulas, diagrams, and pictures. It is known that $ b \le a / 2 $.

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Equation to find $b/a$:
$$
\cos (\pi b /a) = \sqrt{\frac{I(2)}{I(1)}}.
$$
Equation which uses another reflexes are also possible.
Another solutions are possible.
0.40
Measured intensities for two reflexes. 0.20
DG3: $a/b \in [7; 10]$ 0.30
DG4: $a/b \in [3.2; 4.8]$ 0.30
DG5: $a/b =[1.5; 2.5]$ 0.30
A6  0.70 Write down $ \rho (x) $ for the unit cell of the diffraction grating from A1 (Fig.   4A). Use the coordinate system as shown in the figure. Suppose that the unit cell is such that the period of the lattice $ a $ is $ p $ times the width of the slit $ b $: $ a = p b, \quad p \in \mathbb{N} $. Calculate the structure factor $ F_A (h) $ for this unit cell for reflex $ h $. Record your answer using $ h $ and $ q_1 $. What maxima have intensity 0? Write the equation for $ h $ for such maxima.

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$\rho(x) = \begin{cases} 1, & x \in [0,b) \\ 0, & x \in [b, a) \end{cases} $ 0.10
Correct initial integral 0.10
Correct integration 0.20
Final answer
$F_A(h) = 2 \cdot \frac{\sin \left( \pi h/p \right)}{q_1 h} \cdot e^{i \pi h/p } = \begin{cases} \frac{2\pi}{q_1} \cdot \left(\frac{1}{p} \right) , & h = 0 \\
F_1(h), & h \neq 0 \end{cases}$
0.20
$h = \pm pm, \quad m \in \mathbb{N}$ 0.10
A7  0.70 Consider another unit cell (Fig. 4B) of the diffraction grating. Calculate the structure factor $ F_B (h) $ for this unit cell for reflex $ h $. What reflexes of this diffraction grating have an intensity of 0? Write the equation for $ h $ for such reflexes.

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Correct initial integral or idea to subtract. 0.20
Correct answer for $h = 0$
$F_B(0) = \frac{2\pi}{q_1} \cdot \left( 1 - \frac{1}{p} \right)$
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Correct answer for $h\ne0$
$F_B(h) = 2 \cdot \frac{\sin \left( \pi h \right)}{q_1 h} \cdot e^{i \pi h} - 2 \cdot \frac{\sin \left( \pi h/p \right)}{q_1 h} \cdot e^{i \pi h/p } =
\begin{cases} \frac{2\pi}{q_1} \cdot \left( 1 - \frac{1}{p} \right) , & h = 0 \\
- F_1(h), & h \neq 0 \end{cases}$
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$h = \pm pm, \quad m \in \mathbb{N}$ 0.10
A8  0.40 These two diffraction gratings described above are illuminated with light of the same intensity. Find the quotients $ I_{A, h = 0} / I_{B, h = 0} $ and $ I_{A, h = 1} / I_{B, h = 1} $.

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$\frac{I_{A, h=0}}{I_{B,h=0}} = \left( \frac{b}{a-b} \right)^2 = \left( \frac{1}{p-1} \right)^2$
or
$\frac{I_{A, h=0}}{I_{B,h=0}} = \left( \frac{b}{a-b} \right) = \left( \frac{1}{p-1} \right)$
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$\frac{I_{A, h=0}}{I_{B,h=0}} = \left( \frac{b}{a-b} \right)^2 = \left( \frac{1}{p-1} \right)^2$ 0.10
$\frac{I_{A, h=1}}{I_{B,h=1}} = 1$ 0.20
B1  1.00 Find the angle $ \beta $ between the vectors $ \vec{q}_1 $ and $ \vec{q}_2 $ and their lengths $ q_1 $, $ q_2 $. Please note that these vectors must be of minimum length and the angle between them must be $ \le 90 ^ \circ $. Express your answer through the crystal parameters $ a_1, a_2, \alpha $ (Fig.   5)

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$|\vec{q}| \sim \frac{2\pi}{a}$ 0.20
Answer for the first vector
$|\vec{q}_1| = \frac{2\pi}{a_1 \sin \alpha}, \quad |\vec{q}_1| = \frac{2\pi}{a_2 \sin \alpha}$
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Answer for the second vector
$|\vec{q}_2| = \frac{2\pi}{a_2 \sin \alpha}, \quad |\vec{q}_2| = \frac{2\pi}{a_1 \sin \alpha}$
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$\beta = \alpha$ 0.20
B2  1.00 For crystals A and D, find the complex amplitude modulus $ | F (h, k) | $ for the reflex $ (h, k) $. Express your answer in terms of $ a $ (crystal period) and $ b $ (atom size). It is enough to indicate an expression that is true for all reflexes except for the central one ($ h = 0 $, $ k = 0 $)

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Initial integral (it's for opposite unit cell, but for $h\ne0, k\ne0$ it doesn't matter) $$F(h, k) = \int_0^b \int_0^b e^{i q_x x} e^{i q_y y} dx dy$$.
Could be slightly different for two reasons:
-another origin of axis;
-one could use precise integral to direct unit cell
0.20
Correct integration (but may be final answer is not ok) 0.10
$|F(h, k)| = \frac{a^2}{\pi^2} \cdot \left| \frac{\sin (\pi h b/a)}{h} \cdot \frac{\sin (\pi k b/a)}{k} \right|, \quad h\neq0, \quad k\neq0$ 0.20
Correct initial integral OR idea that 2nd atom is shifted:
$F_2(h, k) = F_A(h,k) \cdot e^{i q_x a/2} e^{i q_y a/2}$
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Correct integration (but may be final answer is not ok) 0.10
$|F(h, k)| = \frac{a^2}{\pi^2} \cdot \left| \frac{\sin (\pi h b/a)}{h} \cdot \frac{\sin (\pi k b/a)}{k} \cdot (1+(-1)^{h+k}) \right|, \quad h\neq0, \quad k\neq0$ 0.20
B3  0.60 Look at the diffraction patterns of samples UC1-UC4. For each UC1-UC4 sample, experimentally determine the crystal lattice period $ a_{UC1} $, $ a_{UC2} $, $ a_{UC3} $, $ a_{UC4} $.

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$a_{UC1} = 29.5 \pm 2.0 \mathrm{\mu m}$ 0.15
$a_{UC2} = 19.7 \pm 1.3 \mathrm{\mu m}$ 0.15
$a_{UC3} = 19.7 \pm 1.3 \mathrm{\mu m}$ or $a_{UC3} = 13.9 \pm 1.0 \mathrm{\mu m}$ 0.15
$a_{UC4} = 29.5 \pm 2.0 \mathrm{\mu m}$ 0.15
B4  0.40 For each UC1-UC4 sample, find the corresponding crystal structure among Fig.  6. Explain your choice using diagrams, pictures and formulas.

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1-B 2-A 3-D 4-C
(Enter integer number – number of right matches) (до 4 точек)
4 × 0.10
B5  0.80 Determine the size of the atom $ b $.

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Correct equation to solve. Smth like this:
$$
\cos (\pi b /a) = \sqrt{\frac{I(2)}{I(1)}}.
$$
0.20
Two measured intensities 0.20
Correct root of equation 0.20
$b = 10 \pm 2\mu m$ 0.20
B6  1.20 Observe the diffraction patterns of samples UC5, UC6, UC7. Determine experimentally the parameters $ a_1 $, $ a_2 $ and the angle $ \alpha $ for each sample. Explain which parameters of the diffraction pattern you are using with the help of diagrams and figures.

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UC5: The first one of
$a_1= 19.7 \pm 1.8 \mathrm{\mu m}$, $a_2 = 39.4\pm 3.5 \mathrm{\mu m}$
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UC5: The second one of
$a_1= 19.7 \pm 1.8 \mathrm{\mu m}$, $a_2 = 39.4\pm 3.5 \mathrm{\mu m}$
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UC5: $\alpha = 90 ^\circ$ 0.10
UC6: The first one of
$a_1= 35.5 \pm 3.2 \mathrm{\mu m}$, $a_2 = 22.1 \pm 2.0 \mathrm{\mu m}$

(see also another in solution http://pho.rs/p/115/s)
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UC6: The second one of
$a_1= 35.5 \pm 3.2 \mathrm{\mu m}$, $a_2 = 22.1 \pm 2.0 \mathrm{\mu m}$
0.15
UC6: $\alpha = (63 \pm 6) ^\circ$ 0.10
UC7: The first one of
$a_1= 39.4 \pm 3.5 \mathrm{\mu m}$, $a_2= 35.4 \pm 3.2 \mathrm{\mu m}$

(see also another in solution http://pho.rs/p/115/s)
0.15
UC7: The second one of
$a_1= 39.4 \pm 3.5 \mathrm{\mu m}$, $a_2= 35.4 \pm 3.2 \mathrm{\mu m}$
0.15
UC7: $\alpha = (56 \pm 6) ^\circ$ 0.10
C1  0.30 Specify $ h $ and $ k $ for the point that will be the center of rotation. What orders of rotational symmetry $ m $ are possible for a given image?

Draw all possible axes of mirror symmetry in the image. Name your lines.

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$h = 0$, $k = 0$ 0.10
$m = 1, 2, 4$ 0.10
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C2  0.20 Specify the equation of the straight line for each axis of mirror symmetry drawn in the previous task. Do not forget to note which equation corresponds to which line.

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1: $q_y = 0$ 0.05
2: $q_y = q_x$ 0.05
3: $q_x = 0$ 0.05
4: $q_x = -q_y$ 0.05
C3  0.40 For each rotation symmetry and axis of mirror symmetry, write down the corresponding notation ($ C_m $ for rotation and equation for mirror symmetry) and the equation for the intensities $ I (q_x, q_y) $, which should take place if this symmetry element is present.

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$C_1$: $I(q_x, q_y) = I(q_x, q_y)$ 0.05
$C_2$: $I(q_x, q_y) = I(-q_x, -q_y)$ 0.05
$C_4$: $I(q_x, q_y) = I(-q_y, q_x)$ or $I(q_x, q_y) = I(q_y, -q_x)$ 0.10
$q_y=0$: $I(q_x, q_y) = I(q_x, -q_y)$ 0.05
$q_x=0$: $I(q_x, q_y) = I(-q_x, q_y)$ 0.05
$q_x=q_y$: $I(q_x, q_y) = I(q_y, q_x)$ 0.05
$q_x=-q_y$: $I(q_x, q_y) = I(-q_y, -q_x)$ 0.05
C4  0.20 Write down the equation for the intensities of the reflexes $ (h, k) $ and $ (- h, -k) $. What symmetry from question C1 corresponds to this equation? Explain your answer.

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$I(-h, -k) = I(h, k)$ 0.10
Corresponds to $C_2$ 0.10
C5  0.40 Using the definition of the structure factor and symmetry find the structural factors $ f_2 (q_x, q_y) $, $ f_3 (q_x, q_y) $, $ f_4 (q_x, q_y) $ for crystals 2, 3, 4, respectively. Express your answer in terms of the structure factor $ F (q_x, q_y) = f_1 (q_x, q_y) $ of crystal 1.

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$ f_2 (q_x, q_y) = F(-q_x, q_y)$ 0.10
$ f_3 (q_x, q_y) = F(q_y, q_x)$ 0.10
$ f_4 (q_x, q_y) = F(q_x, q_y) \cdot e^{i(q_x x_1+q_y y_1)}$ 0.20
C6  0.50 Consider an arbitrary 2D crystal (Fig.   5). Indicate what orders of $ m $ symmetry of rotation can be in 2D crystals. Explain the answer.

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Explanation 0.30
$C_1$, $C_2$, $C_3$, $C_4$, $C_6$ 0.20
C7  0.90 Determine what symmetries the crystals with unit cells $ K $, $ L $, $ M $, $ N $, and $ P $, $ Q $, $ R $, $ S $, $ T $ have. (Fig. in the answer sheet). Draw the axes of mirror symmetry, at the bottom of the picture indicate which rotational symmetries are present on it.

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If symmetry $C_1$ is not presented -0.10
C8  0.80 Observe the diffraction patterns of the samples <strong>PG 1, 2, 5, 8</strong>. These samples correspond to the unit cells $ K $, $ L $, $ M $, $ N $. Determine what symmetries the given diffraction patterns have. Find the correspondence between patterns and unit cells.

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0.10
1-L, 2-M, 5-N, 8-K (до 4 точек) 4 × 0.10
C9  1.00 Observe the diffraction of the samples <strong>PG 3, 4, 6, 7, 9</strong>. These samples correspond to the unit cells $ P $, $ Q $, $ R $, $ S $, $ T $. Find the correspondence between samples and unit cells. Explain your solution using formulas, diagrams and pictures.

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$\{4, 6\}$ corresponds to $\{R, T\}$, $\{3, 7, 9\}$ to $\{P, Q, S\}$. 0.25
3-P, 4-R, 6-T, 7-Q, 9-S. (Enter integer number – number of right matches) (до 5 точек) 5 × 0.15
C10  0.30 Observe the diffraction pattern of the UC8 sample. Could this sample be a crystal? Explain your answer.

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Answer - NO 0.10
Explanation – structure has $C_5$ symmetry. 0.20
D1  1.00 The crystal (MR0 or MR2) is illuminated with light with an intensity of $ I_0 $. Find the intensity of the maximum at $ \vec{q} = 0 $.

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Electric field of the reflex $\vec{q} = 0$ is proportional to transparent area of an unit cell and should be $E_0$
$$
E = \frac{N_{transparent}}{N_{all}} E_0.
$$
0.20
$E_{MR0} = \left( \frac{5}{16} \right) \cdot E_0$ 0.30
$I_{MR0} = \left( \frac{5}{16} \right)^2 \cdot I_0$ 0.10
$E_{MR2} = \left( \frac{7}{16} \right) \cdot E_0$ 0.30
$I_{MR2} = \left( \frac{7}{16} \right)^2 \cdot I_0$ 0.10
D2  2.00 Determine the structure of the unit cell of the MR1 crystal. The MR1 crystal has one of the indicated units cell (Fig.  12). Describe your solution.

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Measurements. (Enter integer number – number of right points) (до 25 точек) 25 × 0.02
\begin{equation}
\rho(\chi, \gamma) = \sum_{h=-2}^{2} \sum_{k=-2}^{2} \sqrt{I(h, k)} \cdot \cos \left( \varphi(h, k) - \frac{\pi}{2} (\chi h + \gamma k) \right).
\end{equation}
0.50
Calculations 0.50
Structure X 0.50
D3  2.00 Determine the unit cell structure of the MR2 crystal. The structure of MR2 is similar to MR0: two non-transparent squares have become transparent.

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Measurements. (Enter integer number – number of right points) (до 25 точек) 25 × 0.02
Calculations for each black square. (Enter integer number – number of right points) (до 11 точек) 11 × 0.10
Answer 0.40