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APhO 2026, Теория — Задача 2. Diffraction of X-rays by Structured and Evolving Targets

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Theoretical Exam Q2 — Diffraction of X-rays by Structured and Evolving Targets (10 points) · 10 т.

Условие

Note

Vectors are denoted by bold symbols (e.g., $\mathbf{r}$ , $\mathbf{q}$ ).
Assume that absorption is neglected and that the electric field is polarized perpendicular to the plane of incidence.

X-ray diffraction patterns arise because many tiny "wave sources" within a crystal interfere, and we can predict this by adding their complex amplitudes with the correct phases. Consider a monochromatic wave characterized by a (real) amplitude $A \geq 0$ and a phase $\phi$ . We define the complex amplitude $\tilde{A}$ as

\tilde{A} = Ae^{i\phi},

so that the (real) amplitude of the wave is the magnitude (absolute value) of $\tilde{A}$ , and $\phi$ is its phase. Thus $\tilde{A}$ conveniently encodes both magnitude and phase in a single complex number. In this problem set, we define a dimensionless intensity by the squared magnitude of the total complex amplitude:

I = |\tilde{A}|^2 = A^2.

Common experimental and geometric proportionality factors, such as detector response, incident-beam normalization, and common propagation factors, are absorbed into this definition. By contrast, the single-electron scattering amplitude $f_0$ is retained explicitly as the amplitude scale for scattering from one point electron.

When a crystalline material is exposed to an incident wave, the wave is diffracted by the crystal lattice and the diffracted parts interfere with one another. The intensity of the resulting diffracted wave can be calculated by adding the complex amplitudes of the individual diffracted waves, taking into account the phase differences between them, and then computing the squared magnitude of the resulting total complex amplitude. Diffraction primarily arises from interactions with electrons, and contributions from heavier particles such as nuclei are typically negligible. The amplitude of the wave diffracted by a single point electron depends only on $R = |\mathbf{R}|$ , the distance from the electron to the detector. Since $R$ is much larger than the dimensions of the sample, its variation across the sample can be neglected. Therefore, the total diffracted wave can be accurately determined by properly accounting for the phase differences between individual diffracted waves, while their amplitudes are assumed to be constant.

Let $\mathbf{k}_i$ and $\mathbf{k}_f$ denote the wavevectors of the incident and diffracted waves, respectively. An incident plane wave with wavevector $\mathbf{k}_i$ is diffracted into a wave with wavevector $\mathbf{k}_f$ . The momentum transfer is defined as

\mathbf{q} = \mathbf{k}_f - \mathbf{k}_i

and their magnitudes are assumed to be equal, because the wavelength is unchanged:

k \equiv |\mathbf{k}_i| = |\mathbf{k}_f| = \frac{2\pi}{\lambda}.

A.1 Consider two point electrons located at positions $\mathbf{r}_1$ and $\mathbf{r}_2$ , and define $\mathbf{r} \equiv \mathbf{r}_2 - \mathbf{r}_1$ . A detector is located at $\mathbf{P}$ , and we define $\mathbf{R} \equiv \mathbf{P} - \mathbf{r}_1$ . A plane wave $E_i(\mathbf{r}) \propto e^{i\mathbf{k}_i \cdot \mathbf{r}}$ is incident on the two electrons, and the diffracted wave is observed in the far field along the direction $\mathbf{k}_f$ . In part A, the common time-dependent factor $e^{-i\omega t}$ is suppressed, since only relative spatial phases are relevant.

Under the far-field approximation, $R \equiv |\mathbf{R}| \gg |\mathbf{r}|$ , keep only the leading-order term in $|\mathbf{r}|/R$ and write the outgoing geometric path difference, $\Delta L_{\text{out}} \equiv |\mathbf{P} - \mathbf{r}_1| - |\mathbf{P} - \mathbf{r}_2|$ , in terms of $\mathbf{r}$ and $\mathbf{k}_f$ , or equivalently $\mathbf{k}_f/k_f$ . [0,6 т.]

A.2 Using your result from A.1 and accounting for the position-dependent phase of the incident wave at $\mathbf{r}_1$ and $\mathbf{r}_2$ , find the phase difference between the two diffracted contributions at the detector, expressed in terms of $\mathbf{q}$ and $\mathbf{r}$ . The phase difference is defined as $\Delta\phi \equiv \phi_1 - \phi_2$ , where $\phi_1$ and $\phi_2$ are the phases of the contributions from the electrons at $r_1$ and $r_2$ . [0,4 т.]

A.3 Express the total complex amplitude of the diffracted wave from these two electrons in terms of $\mathbf{q}$ and $\mathbf{r}$ . You may ignore any overall common phase factor, since it does not affect the intensity. Assume that the real amplitude of a diffracted wave from a single point electron is a constant $f_0$ , independent of position. [0,6 т.]

A.4 Express the intensity of diffracted waves from those two electrons in terms of $\mathbf{q}$ and $\mathbf{r}$ . [0,4 т.]

B.1 Consider two point-like electrons located at positions $\mathbf{r}_1$ and $\mathbf{r}_2$ , with $\mathbf{r} \equiv \mathbf{r}_2 - \mathbf{r}_1$ . A beam of wavelength $\lambda_0$ (so $k = 2\pi/\lambda_0$ and $\omega = 2\pi c/\lambda_0$ ) illuminates the electrons, and the diffracted wave is observed in the far field along $\mathbf{k}_f$ . We model the incident field as a plane wave with a time-dependent random phase,

E_i(\mathbf{r}, t) = A\, \exp[i(\mathbf{k}_i \cdot \mathbf{r} - \omega t + \phi(t))],

where $\phi(t)$ is piecewise constant and undergoes random phase jumps at regular time intervals of duration

t_0 \equiv \frac{L_0}{c},

with $L_0$ the (given) longitudinal coherence length. At the beginning of each interval of length $t_0$ , $\phi(t)$ is reset to a new independent value uniformly distributed on $[0, 2\pi)$ . Let $I_0$ denote the (time-averaged) intensity at the detector that would be obtained from a single electron in the same geometry. In this problem, the detector is assumed to measure a time-averaged intensity. That is, it does not resolve the individual random phase jumps. Instead, it records the average of the instantaneous intensity over a time much longer than the phase-jump interval $t_0$ : $\langle I \rangle_t \equiv \langle |E(t)|^2 \rangle_t$ . Here $\langle \cdots \rangle_t$ denotes an average over many phase-jump intervals.

Using the phase-jump model above, derive the time-averaged total intensity $\langle I \rangle_t$ at the detector from the two electrons. Your final result should be written in terms of $I_0$ , $\mathbf{q} \equiv \mathbf{k}_f - \mathbf{k}_i$ , the separation vector $\mathbf{r}$ , the wavelength $\lambda_0$ , and the coherence length $L_0$ . Assume the detector averages over times much longer than $t_0$ . [2,3 т.]

C.1 Part C: Non-point particle effect (1.0 pts)

An electron is often treated as a classical point particle, but in a more realistic model its charge may be regarded as being distributed over a finite spatial region. Consider two idealized charge distributions: (i) an ideal point charge located at $\mathbf{r} = 0$ , whose scattering amplitude is $A_1(\mathbf{q}) = Q_0$ , (ii) an extended Gaussian charge distribution $\rho_2(\mathbf{r}) = \rho_0 \exp\left(-\frac{r^2}{R_0^2}\right)$ , $r = |\mathbf{r}|$ .

The constants $Q_0$ and $\rho_0$ are chosen so that the total charge is the same in both cases:

Q_0 = \int \rho_2(\mathbf{r})\, d^3r.

Here $d^3r$ denotes the volume element in three-dimensional space. In Cartesian coordinates, $d^3r = dx\,dy\,dz$ and $\int_{\mathbb{R}^3}$ means integration over all space. Useful identities (may be used without proof):

\int_0^\infty e^{-r^2/R_0^2}\, 4\pi r^2\, dr = \pi^{3/2} R_0^3, \qquad \int_{\mathbb{R}^3} e^{-\alpha r^2} e^{i\mathbf{k}\cdot\mathbf{r}}\, d^3\mathbf{r} = \left(\frac{\pi}{\alpha}\right)^{3/2} \exp\left(-\frac{k^2}{4\alpha}\right), \qquad \alpha > 0.

Obtain the relation among $Q_0$ , $\rho_0$ , and $R_0$ . [0,4 т.]

C.2 Evaluate the amplitude

A_2(\mathbf{q}) \equiv \int_{\mathbb{R}^3} \rho_2(\mathbf{r}) e^{i\mathbf{q}\cdot\mathbf{r}}\, d^3\mathbf{r}

and compare it with the point-charge amplitude $A_1(\mathbf{q}) = Q_0$ . [0,4 т.]

C.3 Estimate the ratio of the diffracted intensities, $\dfrac{I_2}{I_1}$ , for these two idealized cases when $q = \dfrac{2}{R_0}$ . [0,2 т.]

D.1 Imagine that the surface of a film (i.e., the top atomic layers) is not perfectly flat, but exhibits surface roughness. A common model is to assume that the local film thickness (measured in monolayers) follows a Gaussian distribution. Let $N$ denote the local number of completed monolayers within a lateral coherence area of the beam. We assume that $N$ varies across the surface and is normally distributed with mean $\bar{N}$ and standard deviation $\sigma$ (both in units of monolayers):

P(N) = \frac{1}{\sqrt{2\pi}\,\sigma} \exp\left[-\frac{(N - \bar{N})^2}{2\sigma^2}\right].

(For the purpose of evaluating averages, you may treat $N$ as a continuous variable.) Let $d$ be the spacing between adjacent atomic layers (monolayers), and let $q_z$ denote the component of the scattering vector $\mathbf{q} = \mathbf{k}_f - \mathbf{k}_i$ normal to the flat surface of the film, i.e., $q_z = \mathbf{q} \cdot \hat{\mathbf{z}}$ . For an integer number of monolayers $N$ , the scattering amplitude is

A_N(q_z) = \sum_{n=0}^{N-1} e^{iq_z n d} = \frac{1 - e^{iq_z N d}}{1 - e^{iq_z d}}.

When averaging over the Gaussian thickness distribution, we treat $N$ as a continuous variable and use the closed-form expression

A_N(q_z) \equiv \frac{1 - e^{iq_z Nd}}{1 - e^{iq_z d}}

as the corresponding continuous extension. Assume the measured diffraction intensity is obtained from the coherent average amplitude,

I(q_z) \equiv |\langle A(q_z) \rangle|^2, \qquad \langle A(q_z) \rangle = \int_{-\infty}^{\infty} P(N)\, A_N(q_z)\, dN.

Calculate the intensity ratios

\frac{I(q_z, \sigma = 0.4, \bar{N} = 5)}{I(q_z, \sigma = 0, \bar{N} = 5)}

q_z = \frac{\pi}{2d} \quad \text{and} \quad q_z = \frac{2\pi}{d},

respectively. If needed, evaluate the second case by taking the appropriate limit. [2,4 т.]

E.1 Imagine a thin film with a simple cubic structure grown on a substrate in a layer-by-layer mode, i.e., each monolayer is completed before the next monolayer starts to grow. Let $d$ be the spacing between adjacent atomic layers (monolayers), and let $q_z$ denote the component of $\mathbf{q} = \mathbf{k}_f - \mathbf{k}_i$ normal to the flat surface of the film, i.e., $q_z = \mathbf{q} \cdot \hat{\mathbf{z}}$ . As the film is being grown, the thickness of the film changes, and so does the diffraction intensity at $\mathbf{q} = \frac{\pi}{d}\hat{\mathbf{z}}$ . The calculation is performed at the specified out-of-plane momentum transfer, and all monolayer contributions are coherently summed.

If the film starts to grow at $t = 0$ and the time required to complete one monolayer is $t_0$ , obtain the diffraction intensity ratio

\frac{I(t = 0.8 t_0)}{I(t = 3.6 t_0)}

measured at $\mathbf{q} = \frac{\pi}{d}\hat{\mathbf{z}}$ . Assume that during each monolayer-growth interval, the fractional coverage of the top layer increases linearly from 0 to 1, so that at time $t = (N + \theta)t_0$ , there are $N$ completed monolayers and a fractional coverage $\theta$ of the next monolayer. [2,3 т.]

Решение

General marking instructions

A global phase factor multiplying a complex amplitude is not penalized, because it does not change intensity.
In Part A, an overall sign change in a path or phase difference is acceptable only if the subtraction convention is stated or used consistently.
In Part B, $I_0$ is the single-electron time-averaged intensity. Expressions using $f_0^2$ may receive full credit only if the identification $I_0 = f_0^2$ is explicit or clear from context.

Part A: Diffraction from two electrons treated as point particles [2.0 points]

A.1 [0.6 points]

Target result:

\Delta L_{\mathrm{out}} = |\mathbf{P} - \mathbf{r}_1| - |\mathbf{P} - \mathbf{r}_2| \simeq \mathbf{r} \cdot \frac{\mathbf{k}_f}{k_f}.

Award marks only for the following identifiable equations or equivalent expressions:

(a) States the reference distance

|\mathbf{P} - \mathbf{r}_1| = |\mathbf{R}| = R.

[0.10 pt]

(b) Rewrites the second path length using the separation vector

|\mathbf{P} - \mathbf{r}_2| = |\mathbf{R} - (\mathbf{r}_2 - \mathbf{r}_1)| = |\mathbf{R} - \mathbf{r}|.

[0.10 pt]

|\mathbf{R} - \mathbf{r}| \simeq R - \hat{\mathbf{s}}_f \cdot \mathbf{r} \quad \text{with} \quad \hat{\mathbf{s}}_f = \frac{\mathbf{R}}{R}.

[0.20 pt]

(d) Identifies the far-field detector direction with the outgoing wavevector direction

\hat{\mathbf{s}}_f \simeq \frac{\mathbf{k}_f}{k_f}.

[0.10 pt]

(e) Obtains the final path difference

\Delta L_{\mathrm{out}} \simeq \hat{\mathbf{s}}_f \cdot \mathbf{r} = \mathbf{r} \cdot \frac{\mathbf{k}_f}{k_f}.

[0.10 pt]

If the student defines the path difference in the reverse order and obtains $-\mathbf{r} \cdot \mathbf{k}_f/k_f$ , award full credit if the convention is stated or used consistently.

A.2 [0.4 points]

Target result:

\Delta \phi = \mathbf{q} \cdot \mathbf{r}.

Award marks only for the following identifiable equations or equivalent expressions:

(a) Writes the incident-wave phase difference in the chosen convention, for example

\Delta \phi_{\mathrm{in}} = \mathbf{k}_i \cdot \mathbf{r}_1 - \mathbf{k}_i \cdot \mathbf{r}_2 = -\mathbf{k}_i \cdot \mathbf{r}.

[0.10 pt]

(b) Writes the outgoing propagation phase contribution using A.1,

\Delta \phi_{\mathrm{out}} = k\Delta L_{\mathrm{out}} = k\, \mathbf{r} \cdot \frac{\mathbf{k}_f}{k_f} = \mathbf{k}_f \cdot \mathbf{r}.

[0.10 pt]

\Delta \phi = \Delta \phi_{\mathrm{in}} + \Delta \phi_{\mathrm{out}} = (-\mathbf{k}_i + \mathbf{k}_f) \cdot \mathbf{r}.

[0.10 pt]

(d) Uses $\mathbf{q} = \mathbf{k}_f - \mathbf{k}_i$ to obtain

\Delta \phi = (\mathbf{k}_f - \mathbf{k}_i) \cdot \mathbf{r} = \mathbf{q} \cdot \mathbf{r}.

[0.10 pt]

Accept $-\mathbf{q} \cdot \mathbf{r}$ for full credit only if the student has explicitly adopted the opposite phase-difference convention.

A.3 [0.6 points]

Target result, up to a global phase:

\widetilde{A}_{\mathrm{tot}} = f_0 \left(1 + e^{i\mathbf{q}\mathbf{r}}\right).

Award marks only for the following identifiable equations or equivalent expressions:

(a) Writes a valid single-electron amplitude, e.g.

\widetilde{A}_n = f_0 e^{-i\mathbf{q}\cdot\mathbf{r}_n} \quad \text{or} \quad \widetilde{A}_n = f_0 e^{+i\mathbf{q}\cdot\mathbf{r}_n}

with a consistent sign convention. [0.20 pt]

(b) Sums the two electron contributions, e.g.

\widetilde{A}_{\mathrm{tot}} = f_0 e^{-i\mathbf{q}\cdot\mathbf{r}_1} + f_0 e^{-i\mathbf{q}\cdot\mathbf{r}_2}.

[0.20 pt]

\widetilde{A}_{\mathrm{tot}} = f_0 e^{-i\mathbf{q}\cdot\mathbf{r}_2} \left(1 + e^{i\mathbf{q}\cdot(\mathbf{r}_2-\mathbf{r}_1)}\right) = f_0 e^{-i\mathbf{q}\cdot\mathbf{r}_2} \left(1 + e^{i\mathbf{q}\mathbf{r}}\right).

[0.10 pt]

(d) Gives a final equivalent form after dropping the global phase, for example

\widetilde{A}_{\mathrm{tot}} = f_0(1 + e^{i\mathbf{q}\mathbf{r}}) \quad \text{or} \quad \widetilde{A}_{\mathrm{tot}} = 2f_0 e^{i\mathbf{q}\mathbf{r}/2} \cos\left(\frac{\mathbf{q}\cdot\mathbf{r}}{2}\right).

[0.10 pt]

The complex conjugate form $f_0(1 + e^{-i\mathbf{q}\mathbf{r}})$ receives full credit because it gives the same intensity.

A.4 [0.4 points]

Target result:

I(\mathbf{q}, \mathbf{r}) = 2f_0^2[1 + \cos(\mathbf{q}\cdot\mathbf{r})] = 4f_0^2 \cos^2\left(\frac{\mathbf{q}\cdot\mathbf{r}}{2}\right).

Award marks only for the following identifiable equations or equivalent expressions:

(a) Uses the intensity definition

I = |\widetilde{A}_{\mathrm{tot}}|^2 = f_0^2 |1 + e^{i\mathbf{q}\mathbf{r}}|^2.

[0.10 pt]

(b) Evaluates the modulus squared correctly,

|1 + e^{ix}|^2 = (1+e^{ix})(1+e^{-ix}) = 2 + 2\cos x.

[0.20 pt]

I = 2f_0^2[1+\cos(\mathbf{q}\cdot\mathbf{r})] \quad \text{or} \quad I = 4f_0^2 \cos^2(\mathbf{q}\cdot\mathbf{r}/2).

[0.10 pt]

Part B: Finite longitudinal coherence (phase-jump model) [2.3 points]

Target result:

\langle I(\mathbf{q}, \mathbf{r}, \lambda_0, L_0) \rangle_t = 2I_0 [1 + \mu(\mathbf{q}, \mathbf{r}; L_0) \cos(\mathbf{q}\cdot\mathbf{r})],

where

\mu(\mathbf{q}, \mathbf{r}; L_0) = \begin{cases} 1 - \dfrac{\lambda_0}{2\pi L_0}|\mathbf{q}\cdot\mathbf{r}|, & |\mathbf{q}\cdot\mathbf{r}| \le \dfrac{2\pi L_0}{\lambda_0}, \\ 0, & |\mathbf{q}\cdot\mathbf{r}| > \dfrac{2\pi L_0}{\lambda_0}. \end{cases}

Equivalently,

\langle I \rangle_t = \begin{cases} 2I_0\left[1 + \left(1 - \dfrac{\lambda_0}{2\pi L_0}|\mathbf{q}\cdot\mathbf{r}|\right)\cos(\mathbf{q}\cdot\mathbf{r})\right], & |\mathbf{q}\cdot\mathbf{r}| \le \dfrac{2\pi L_0}{\lambda_0}, \\ 2I_0, & |\mathbf{q}\cdot\mathbf{r}| > \dfrac{2\pi L_0}{\lambda_0}. \end{cases}

Award marks only for the following identifiable equations or equivalent expressions:

(a) Sets up the two-field total field and time average:

E(t) = E_1(t) + E_2(t), \qquad \langle I \rangle_t = \langle |E_1(t)+E_2(t)|^2 \rangle_t.

[0.20 pt]

(b) Expands the intensity and identifies the interference term:

|E_1+E_2|^2 = |E_1|^2 + |E_2|^2 + 2\Re(E_1E_2^*),

\langle I \rangle_t = 2I_0 + 2\Re\langle E_1E_2^*\rangle_t.

[0.30 pt]

\langle E_1E_2^*\rangle_t = E_0^2 \langle e^{i[\phi_1(t)-\phi_2(t)]}\rangle_t.

[0.25 pt]

(d) Separates deterministic and random phases and identifies the deterministic part:

\phi_1(t) - \phi_2(t) = \Delta\phi + \delta\phi(t), \qquad \Delta\phi = \mathbf{q}\cdot\mathbf{r}.

[0.30 pt]

(e) Defines the coherence factor explicitly as

\mu(\Delta t) = \langle e^{i\delta\phi(t)}\rangle_t, \qquad \langle E_1E_2^*\rangle_t = E_0^2 \mu(\Delta t) e^{i\Delta\phi}.

[0.20 pt]

(f) Writes the same-interval probability for the phase-jump model:

P_{\mathrm{same}}(\Delta t) = \begin{cases} 1 - |\Delta t|/t_0, & |\Delta t| < t_0, \\ 0, & |\Delta t| \ge t_0. \end{cases}

where

P_{\mathrm{same}}(\Delta t) \equiv \Pr\{t \text{ and } t+\Delta t \text{ lie in the same interval of length } t_0\}.

[0.35 pt]

(g) Shows the averaging rule for the random phase:

e^{i\delta\phi} = 1 \quad \text{for same interval}, \qquad \langle e^{i\delta\phi}\rangle = 0 \quad \text{for different intervals},

and hence

\mu(\Delta t) = P_{\mathrm{same}}(\Delta t).

[0.25 pt]

(h) Relates time delay to path/phase difference:

\Delta\phi = \frac{2\pi}{\lambda_0}\Delta\ell = \mathbf{q}\cdot\mathbf{r}, \qquad |\Delta t| = \frac{|\Delta\ell|}{c} = \frac{\lambda_0}{2\pi c}|\mathbf{q}\cdot\mathbf{r}|, \qquad t_0 = \frac{L_0}{c}.

[0.20 pt]

(i) Gives the correct final intensity formula including the no-interference limit:

\langle I \rangle_t = 2I_0[1 + \mu\cos(\mathbf{q}\cdot\mathbf{r})],

with

\mu = 1 - \frac{\lambda_0}{2\pi L_0}|\mathbf{q}\cdot\mathbf{r}| \text{ inside the coherence window}, \qquad \langle I \rangle_t = 2I_0 \text{ outside it}.

[0.25 pt]

The random phase $\phi(t)$ represents the source phase carried by the incident wave. Therefore, at position $\mathbf{r}$ , it should be understood as being evaluated at the retarded time $t - \hat{\mathbf{s}}_i \cdot \mathbf{r}/c$ , where $\hat{\mathbf{s}}_i = \mathbf{k}_i/k$ . Consequently, the two scattered contributions sample the random source phase at times separated by the total effective path delay corresponding to the full scattering phase difference $\Delta\phi = \mathbf{q}\cdot\mathbf{r}$ , not only by the outgoing path difference.

Part C: Non-point-particle effect [1.0 point]

C.1 [0.4 points]

Target result:

Q_0 = \rho_0 \pi^{3/2} R_0^3, \qquad \rho_0 = \frac{Q_0}{\pi^{3/2}R_0^3}.

Award marks only for the following identifiable equations or equivalent expressions:

(a) States or uses the total-charge equality for the extended distribution:

Q_0 = \int \rho_2(\mathbf{r})\, d^3\mathbf{r}.

[0.10 pt]

(b) Writes the Gaussian total-charge integral in spherical coordinates:

\int \rho_2(\mathbf{r})\, d^3\mathbf{r} = \rho_0 \int_0^\infty e^{-r^2/R_0^2} 4\pi r^2\, dr.

[0.10 pt]

\int \rho_2(\mathbf{r})\, d^3\mathbf{r} = \rho_0 \pi^{3/2} R_0^3.

[0.10 pt]

(d) Gives the normalization relation

Q_0 = \rho_0 \pi^{3/2} R_0^3 \quad \text{or} \quad \rho_0 = \frac{Q_0}{\pi^{3/2}R_0^3}.

[0.10 pt]

C.2 [0.4 points]

Target results:

A_2(\mathbf{q}) = Q_0 \exp\left(-\frac{q^2R_0^2}{4}\right), \qquad \frac{A_2(\mathbf{q})}{A_1(\mathbf{q})} = \exp\left(-\frac{q^2R_0^2}{4}\right),

where the problem gives the point-charge amplitude $A_1(\mathbf{q}) = Q_0$ . Award marks only for the following identifiable equations or equivalent expressions:

(a) Sets up the Gaussian amplitude integral and applies the supplied Fourier-transform identity with $\alpha = 1/R_0^2$ and $k = q$ :

A_2(\mathbf{q}) = \rho_0 \int_{\mathbb{R}^3} e^{-r^2/R_0^2} e^{i\mathbf{q}\cdot\mathbf{r}}\, d^3\mathbf{r}.

[0.10 pt]

(b) Obtains the Gaussian Fourier-transform factor:

\int_{\mathbb{R}^3} e^{-r^2/R_0^2} e^{i\mathbf{q}\cdot\mathbf{r}}\, d^3\mathbf{r} = \pi^{3/2}R_0^3 \exp\left(-\frac{q^2R_0^2}{4}\right).

[0.12 pt]

A_2(\mathbf{q}) = Q_0 \exp\left(-\frac{q^2R_0^2}{4}\right).

[0.13 pt]

(d) Compares with the point-charge amplitude $A_1(\mathbf{q}) = Q_0$ , for example by writing

\frac{A_2(\mathbf{q})}{A_1(\mathbf{q})} = \exp\left(-\frac{q^2R_0^2}{4}\right),

or by stating that the extended-charge amplitude is reduced at nonzero $q$ while matching the point-charge amplitude at $q=0$ . [0.05 pt]

If the student derives or states $A_1(\mathbf{q}) = Q_0$ again, do not penalize; however, no separate credit is required for deriving $A_1$ , because it is given in the problem statement.

C.3 [0.2 points]

Target result:

\frac{I_2}{I_1} = e^{-2} \simeq 0.14.

Award marks only for the following identifiable equations or equivalent expressions:

(a) Squares the amplitudes and obtains

\frac{I_2}{I_1} = \left|\frac{A_2}{A_1}\right|^2 = \exp\left(-\frac{q^2R_0^2}{2}\right).

[0.10 pt]

(b) Substitutes $q = 2/R_0$ and obtains

\frac{I_2}{I_1} = \exp\left[-\frac{(2/R_0)^2 R_0^2}{2}\right] = e^{-2} \simeq 0.14.

[0.10 pt]

If the student gives the amplitude ratio $e^{-1}$ instead of the intensity ratio, award 0.10.

Part D: Diffraction from a film with non-flat surface morphology [2.4 points]

Target results:

\frac{I(q_z=\pi/2d, \sigma=0.4, \bar N=5)}{I(q_z=\pi/2d, \sigma=0, \bar N=5)} = \frac{1+e^{-\pi^2/25}}{2} \simeq 0.84,

\frac{I(q_z=2\pi/d, \sigma=0.4, \bar N=5)}{I(q_z=2\pi/d, \sigma=0, \bar N=5)} = 1.

Award marks only for the following identifiable equations or equivalent expressions:

(a) Writes the column amplitude as a geometric sum and closed form:

A_N(q_z) = \sum_{n=0}^{N-1} e^{iq_z nd} = \frac{1-e^{iq_zNd}}{1-e^{iq_zd}}.

[0.25 pt]

(b) Uses the coherent average, not the incoherent average:

\langle A(q_z)\rangle = \int P(N)A_N(q_z)\, dN, \qquad I(q_z) = |\langle A(q_z)\rangle|^2.

[0.25 pt]

\langle A(q_z)\rangle = \frac{1-\langle e^{iq_zNd}\rangle}{1-e^{iq_zd}}.

[0.20 pt]

(d) Decomposes the scattering vector near a Bragg point:

q_z = Q_z + \Delta q_z, \qquad Q_z = \frac{2\pi m}{d},

and, for integer $N$ , uses

e^{iQ_zNd} = e^{i2\pi mN} = 1,

so that

\langle e^{iq_zNd}\rangle = \langle e^{i\Delta q_zNd}\rangle.

[Reason for using $\Delta q_z$ rather than $q_z$ : For an integer number of layers, writing $q_z = Q_z + \Delta q_z$ with $Q_z = 2\pi m/d$ gives

e^{iq_zNd} = e^{iQ_zNd}e^{i\Delta q_zNd} = e^{i2\pi mN}e^{i\Delta q_zNd} = e^{i\Delta q_zNd}.

Thus, in the underlying integer-layer model, the Bragg-periodic part contributes only a trivial phase factor, and the thickness-dependent dephasing is governed by the deviation $\Delta q_z$ from the Bragg point. In the present continuous-extension model for Gaussian roughness, we apply the Gaussian characteristic function to this residual phase $\Delta q_zNd$ . Therefore, the damping factor is written in terms of $(\Delta q_zd)^2\sigma^2$ , not $(q_zd)^2\sigma^2$ as shown in (e).]
[0.25 pt]

(e) Evaluates the Gaussian characteristic function:

\langle e^{i\Delta q_zNd}\rangle = \int_{-\infty}^{\infty} P(N) e^{i\Delta q_zNd}\, dN.

[0.08 pt]

With $\xi = N - \bar N$ , i.e. $N = \bar N + \xi$ and $dN = d\xi$ ,

\langle e^{i\Delta q_zNd}\rangle = e^{i\Delta q_z\bar Nd} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}\sigma} \exp\left[-\frac{\xi^2}{2\sigma^2}+i\Delta q_zd\,\xi\right] d\xi.

[0.10 pt]

Completes the square correctly:

-\frac{\xi^2}{2\sigma^2}+i\Delta q_zd\,\xi = -\frac{(\xi-i\sigma^2\Delta q_zd)^2}{2\sigma^2} - \frac{1}{2}(\Delta q_zd)^2\sigma^2.

[0.12 pt]

Therefore obtains the exact Gaussian characteristic function

\langle e^{i\Delta q_zNd}\rangle = e^{i\Delta q_z\bar Nd}e^{-\frac{1}{2}(\Delta q_zd)^2\sigma^2}.

This Gaussian integral is exact for the assumed continuous Gaussian distribution; the use of $\Delta q_z$ reflects the Bragg-periodic phase removal described in part (d), not a small- $\sigma$ approximation.
[0.10 pt]

(f) Obtains the averaged amplitude:

\langle A(q_z)\rangle = \frac{1-e^{i\Delta q_z\bar Nd}e^{-\frac{1}{2}(\Delta q_zd)^2\sigma^2}}{1-e^{iq_zd}}.

[0.20 pt]

(g) Computes the coherent intensity:

I(q_z) = \frac{1-2e^{-\frac{1}{2}(\Delta q_zd)^2\sigma^2}\cos(\Delta q_z\bar Nd)+e^{-(\Delta q_zd)^2\sigma^2}}{4\sin^2(q_zd/2)}.

[0.30 pt]

(h) For $q_z = \pi/(2d)$ , identifies the required substitutions:

\Delta q_z d = \frac{\pi}{2}, \qquad \Delta q_z \bar N d = \frac{5\pi}{2}, \qquad \cos\left(\frac{5\pi}{2}\right) = 0, \qquad \sigma^2 = (0.4)^2 = 0.16.

[0.10 pt]

(i) Obtains the first intensity ratio:

\frac{I(\sigma=0.4)}{I(\sigma=0)} = \frac{1 + e^{-(\pi/2)^2(0.4)^2}}{2} = \frac{1 + e^{-\pi^2/25}}{2} \simeq 0.84.

[0.20 pt]

(j) For $q_z = 2\pi/d$ , recognizes the limiting form:

q_z = \frac{2\pi}{d} \Rightarrow \Delta q_z = 0, \qquad \lim_{q_z \to 2\pi/d} A_N(q_z) = N.

[0.15 pt]

(k) Obtains the second requested ratio:

\frac{I(q_z = 2\pi/d, \sigma=0.4, \bar N = 5)}{I(q_z = 2\pi/d, \sigma=0, \bar N = 5)} = 1.

[0.10 pt]

Part E: Diffraction from a film with evolving surface morphology [2.3 points]

Target result:

\frac{I(0.8t_0)}{I(3.6t_0)} = 4.

Award marks only for the following identifiable equations or equivalent expressions:

(a) Identifies the alternating layer phase at $\mathbf{q} = (\pi/d)\hat{\mathbf{z}}$ :

e^{iq_z n d} = e^{i(\pi/d)nd} = e^{in\pi} = (-1)^n.

[0.40 pt]

(b) Writes the amplitude for $N$ completed layers plus fractional coverage $\theta$ :

A(N,\theta) = f_0\left[\sum_{n=0}^{N-1}(-1)^n + \theta(-1)^N\right].

[0.50 pt]

t = 0.8t_0: \quad N=0, \quad \theta=0.8,

t = 3.6t_0: \quad N=3, \quad \theta=0.6.

[0.40 pt]

(d) Computes the first amplitude and intensity:

A(0,0.8) = 0.8 f_0, \qquad I(0.8t_0) = |0.8f_0|^2 = 0.64 f_0^2.

[0.35 pt]

(e) Computes the second amplitude and intensity:

\sum_{n=0}^{2}(-1)^n = 1, \qquad A(3,0.6) = f_0[1 + 0.6(-1)^3] = 0.4 f_0,

I(3.6t_0) = |0.4f_0|^2 = 0.16 f_0^2.

[0.35 pt]

(f) Gives the final ratio:

\frac{I(0.8t_0)}{I(3.6t_0)} = \frac{0.64 f_0^2}{0.16 f_0^2} = 4.

[0.30 pt]

Оригинал в Архива: T2.pdf · официални решения: T2_sol.pdf