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Crystallography

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

1 $
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
$
0.30
A2  0.20 Find scattering vector $q$ for the maximum numbered $h$ for a diffraction grating with a period $a$.

1 $q = \frac{2\pi}{a} \cdot h, \quad h \in \mathbb{Z}$

DISCLAIMER. Here and below we use the notations:
$\mathbb{Z}$ – integer numbers: $\{..., -2, -1, 0, 1, 2, ...\}$
$\mathbb{N}$ – positive numbers: $\{1, 2, ...\}$
0.20
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?

1 $q = q_1 \cdot h, \quad h \in \mathbb{Z}$ 0.10
2 $q_1 \cdot a = 2\pi$ 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.

1 Scheme of setup 0.10
2 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
3 DG1: $ q_1 = 315 \pm 22 \mathrm{~mm^{-1}} $ 0.08
4 DG1: $a = 19.7 \pm 1.3 \mathrm{~\mu m} $ 0.08
5 DG2: $ q_1 = 125 \pm 10 \mathrm{~mm^{-1}} $ 0.08
6 DG2: $ a = 49.2 \pm 3.2 \mathrm{~\mu m} $ 0.08
7 DG3: $ q_1 = 78.5 \pm 5.5 \mathrm{~mm^{-1}} $ 0.08
8 DG3: $a = 78.8 \pm 5.2 \mathrm{~\mu m} $ 0.08
9 DG4: $ q_1 = 78.5 \pm 5.5 \mathrm{~mm^{-1}} $ 0.08
10 DG4: $a = 78.8 \pm 5.2 \mathrm{~\mu m} $ 0.08
11 DG5: $ q_1 = 78.5 \pm 5.5 \mathrm{~mm^{-1}} $ 0.08
12 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 $.

1 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
2 Measured intensities for two reflexes. 0.20
3 DG3: $a/b \in [7; 10]$ 0.30
4 DG4: $a/b \in [3.2; 4.8]$ 0.30
5 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.

1 $\rho(x) = \begin{cases} 1, & x \in [0,b) \\ 0, & x \in [b, a) \end{cases} $ 0.10
2 Correct initial integral 0.10
3 Correct integration 0.20
4 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
5 $h = \pm pm, \quad m \in \mathbb{N}$ 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.

2 Correct initial integral or idea to subtract. 0.20
3 Correct answer for $h = 0$
$F_B(0) = \frac{2\pi}{q_1} \cdot \left( 1 - \frac{1}{p} \right)$
0.20
4 Correct answer for $h\ne0$
$F_B(h) = 2 \cdot \frac{\sin \left( \pi h \right)}{q_1 h} \cdot e^{i \pi h} - 2 \cdot \frac{\sin \left( \pi h/p \right)}{q_1 h} \cdot e^{i \pi h/p } =
\begin{cases} \frac{2\pi}{q_1} \cdot \left( 1 - \frac{1}{p} \right) , & h = 0 \\
- F_1(h), & h \neq 0 \end{cases}$
0.20
5 $h = \pm pm, \quad m \in \mathbb{N}$ 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} $.

1 $\frac{I_{A, h=0}}{I_{B,h=0}} = \left( \frac{b}{a-b} \right)^2 = \left( \frac{1}{p-1} \right)^2$
or
$\frac{I_{A, h=0}}{I_{B,h=0}} = \left( \frac{b}{a-b} \right) = \left( \frac{1}{p-1} \right)$
0.10
2 $\frac{I_{A, h=0}}{I_{B,h=0}} = \left( \frac{b}{a-b} \right)^2 = \left( \frac{1}{p-1} \right)^2$ 0.10
3 $\frac{I_{A, h=1}}{I_{B,h=1}} = 1$ 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)

1 $|\vec{q}| \sim \frac{2\pi}{a}$ 0.20
2 Answer for the first vector
$|\vec{q}_1| = \frac{2\pi}{a_1 \sin \alpha}, \quad |\vec{q}_1| = \frac{2\pi}{a_2 \sin \alpha}$
0.30
3 Answer for the second vector
$|\vec{q}_2| = \frac{2\pi}{a_2 \sin \alpha}, \quad |\vec{q}_2| = \frac{2\pi}{a_1 \sin \alpha}$
0.30
5 $\beta = \alpha$ 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 $)

1 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
2 Correct integration (but may be final answer is not ok) 0.10
3 $|F(h, k)| = \frac{a^2}{\pi^2} \cdot \left| \frac{\sin (\pi h b/a)}{h} \cdot \frac{\sin (\pi k b/a)}{k} \right|, \quad h\neq0, \quad k\neq0$ 0.20
4 Correct initial integral OR idea that 2nd atom is shifted:
$F_2(h, k) = F_A(h,k) \cdot e^{i q_x a/2} e^{i q_y a/2}$
0.20
5 Correct integration (but may be final answer is not ok) 0.10
6 $|F(h, k)| = \frac{a^2}{\pi^2} \cdot \left| \frac{\sin (\pi h b/a)}{h} \cdot \frac{\sin (\pi k b/a)}{k} \cdot (1+(-1)^{h+k}) \right|, \quad h\neq0, \quad k\neq0$ 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} $.

1 $a_{UC1} = 29.5 \pm 2.0~\mathrm{\mu m}$ 0.15
2 $a_{UC2} = 19.7 \pm 1.3~\mathrm{\mu m}$ 0.15
3 $a_{UC3} = 19.7 \pm 1.3~\mathrm{\mu m}$ or $a_{UC3} = 13.9 \pm 1.0~\mathrm{\mu m}$ 0.15
4 $a_{UC4} = 29.5 \pm 2.0~\mathrm{\mu m}$ 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.

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

2 Correct equation to solve. Smth like this:
$$
\cos (\pi b /a) = \sqrt{\frac{I(2)}{I(1)}}.
$$
0.20
3 Two measured intensities 0.20
4 Correct root of equation 0.20
5 $b = 10 \pm 2\mu m$ 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.

1 UC5: The first one of
$a_1= 19.7 \pm 1.8~\mathrm{\mu m}$, $a_2 = 39.4\pm 3.5~\mathrm{\mu m}$
0.15
2 UC5: The second one of
$a_1= 19.7 \pm 1.8~\mathrm{\mu m}$, $a_2 = 39.4\pm 3.5~\mathrm{\mu m}$
0.15
3 UC5: $\alpha = 90 ^\circ$ 0.10
4 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)
0.15
5 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
6 UC6: $\alpha = (63 \pm 6) ^\circ$ 0.10
8 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
9 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
10 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.

1 $h = 0$, $k = 0$ 0.10
2 $m = 1,~2,~4$ 0.10
3 0.10
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.

1 1: $q_y = 0$ 0.05
2 2: $q_y = q_x$ 0.05
3 3: $q_x = 0$ 0.05
4 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.

1 $C_1$: $I(q_x, q_y) = I(q_x, q_y)$ 0.05
2 $C_2$: $I(q_x, q_y) = I(-q_x, -q_y)$ 0.05
3 $C_4$: $I(q_x, q_y) = I(-q_y, q_x)$ or $I(q_x, q_y) = I(q_y, -q_x)$ 0.10
4 $q_y=0$: $I(q_x, q_y) = I(q_x, -q_y)$ 0.05
5 $q_x=0$: $I(q_x, q_y) = I(-q_x, q_y)$ 0.05
6 $q_x=q_y$: $I(q_x, q_y) = I(q_y, q_x)$ 0.05
7 $q_x=-q_y$: $I(q_x, q_y) = I(-q_y, -q_x)$ 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.

1 $I(-h, -k) = I(h, k)$ 0.10
2 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.

1 $ f_2 (q_x, q_y) = F(-q_x, q_y)$ 0.10
2 $ f_3 (q_x, q_y) = F(q_y, q_x)$ 0.10
3 $ 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.

1 Explanation 0.30
2 $C_1$, $C_2$, $C_3$, $C_4$, $C_6$ 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.

1 0.10
2 0.10
3 0.10
4 0.10
5 0.10
6 0.10
7 0.10
8 0.10
9 0.10
10 If symmetry $C_1$ is not presented -0.10
C8  0.80 Observe the diffraction patterns of the samples PG 1, 2, 5, 8. 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.

1 0.10
2 0.10
3 0.10
4 0.10
5 1-L, 2-M, 5-N, 8-K 4 × 0.10
C9  1.00 Observe the diffraction of the samples PG 3, 4, 6, 7, 9. 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.

1 $\{4, 6\}$ corresponds to $\{R, T\}$, $\{3, 7, 9\}$ to $\{P, Q, S\}$. 0.25
2 3-P, 4-R, 6-T, 7-Q, 9-S. (Enter integer number – number of right matches) 5 × 0.15
C10  0.30 Observe the diffraction pattern of the UC8 sample. Could this sample be a crystal? Explain your answer.

1 Answer - NO 0.10
2 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 $.

1 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
2 $E_{MR0} = \left( \frac{5}{16} \right) \cdot E_0$ 0.30
3 $I_{MR0} = \left( \frac{5}{16} \right)^2 \cdot I_0$ 0.10
4 $E_{MR2} = \left( \frac{7}{16} \right) \cdot E_0$ 0.30
5 $I_{MR2} = \left( \frac{7}{16} \right)^2 \cdot I_0$ 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.

1 Measurements. (Enter integer number – number of right points) 25 × 0.02
2 \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
3 Calculations 0.50
4 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.

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