Съдържание

APhO 2001, Теория — Задача 3. Thermal Vibrations of Surface Atoms

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от APhO 2001

APHO II 2001 Theoretical Question 3 — Thermal Vibrations of Surface Atoms

Условие

This question considers the thermal vibrations of surface atoms in an elemental metallic crystal with a face-centered cubic (fcc) lattice structure. The unit cubic cell of an fcc lattice consists of one atom at each corner and one atom at the center of each face of the cubic cell, as shown in Fig. 3a. For the crystal under consideration, we use ( $a$ , 0, 0), (0, $a$ , 0) and (0, 0, $a$ ) to represent the locations of the three atoms on the $x$ , $y$ and $z$ axes of its cell. The lattice constant $a$ is equal to 3.92 Å (i.e..the length of each side of the cube is 3.92 Å).

Unit cubic cell of an fcc lattice with axes x, y, z. Corner atoms labelled A, B, C, D, E, F, G, with G at (0,0, a), E at (0, a, 0) and B at ( a,0,0); face-centre atoms shown as filled and open circles joined by dashed lines to the corners. — Fig. 3a

(1) (a) (1) The crystal is cut in such a way that the plane containing ABCD becomes a boundary surface and is chosen for doing low-energy electron diffraction experiments. A collimated beam of electrons with kinetic energy of 64.0 eV is incident on this surface plane at an incident angle $\phi_0$ of 15.0 $^{\mathrm{o}}$ . Note that $\phi_0$ is the angle between the incident electron beam and the normal of the surface plane. The plane containing $\overline{AC}$ and the normal of the surface plane is the plane of incidence. For simplicity, we assume that all incident electrons are back scattered only by the surface atoms on the topmost layer.

(a) What is the wavelength of the matter waves of the incident electrons?

(1) (b) If a detector is set up to detect electrons that do not leave the plane of incidence after being diffracted, at what angles with the normal of the surface will these diffracted electrons be observable?

(2) (a) (2) Assume that the thermal vibrational motions of the surface atoms are simple harmonic. The amplitude of vibration increases as the temperature rises. Low-energy electron diffraction provides a way to measure the average amplitude of vibration. The intensity $I$ of the diffracted beam is proportional to the number of scattered electrons per second. The relation between the intensity $I$ and the displacement $\vec{u}$ (t) of the surface atoms is given by

I = I_0\ \exp\ \{-\lt [(\vec{K}' - \vec{K})\cdot\vec{u}]^2 \gt\} \qquad (1)

In Eq.(1), $I$ and $I_0$ are the intensities at temperature $T$ and absolute zero, respectively. $\vec{K}$ and $\vec{K}'$ are wave vectors of incident electron and diffracted electron, respectively. The angle brackets < > is used to denote average over time. Note that the relation between the wave vector $\vec{K}$ and the momentum $\vec{p}$ of a particle is $\vec{K} = 2\pi\ \vec{p}\,/\,h$ , where $h$ is the Planck constant.

To measure vibration amplitudes of surface atoms of a metallic crystal, a collimated electron beam with kinetic energy of 64.0 eV is incident on a crystal surface at an incident angle of 15.0 $^{\mathrm{o}}$ . The detector is set up for measuring specularly reflected electrons. Only elastically scattered electrons are detected. A plot of ln ( $I$ / $I_0$ ) versus temperature $T$ is shown in Fig. 3b.

Assume the total energy of an atom vibrating in the direction of the surface normal $\hat{x}$ is given by $k_BT$ , where $k_B$ is the Boltzmann constant.

(a) Calculate the frequency of vibration in the direction of the surface normal for the surface atoms.

(2) (b) Calculate the root-mean-square displacement, i. e. the value of $(\lt u_{\mathrm{x}}^{2}\gt)^{1/2}$ , in the direction of the surface normal for the surface atoms at 300 K.

Graph of ln(I/I0) (vertical axis, 0 to -1.0 in steps of 0.2) versus T (K) (horizontal axis, 0 to 400 in steps of 50) with data points from about 100 K to 350 K and a straight fitted line falling from about -0.22 at 100 K to about -0.85 at 375 K. — Fig. 3b

The following data are given:

Atomic weight of the metal $M$ = 195.1

Boltzmann constant $k_B = 1.38\times10^{-23}\ \mathrm{J/K}$

Mass of electron = $9.11\times10^{-31}\ \mathrm{kg}$

Charge of electron = $1.60\times10^{-19}\ \mathrm{C}$

Planck constant $h = 6.63\times10^{-34}\ \mathrm{J\text{-}s}$

Отговори

Решение

(1) (a) The wavelength of the incident electron is

\begin{aligned}\lambda &= \frac{h}{p} = \frac{h}{\sqrt{2meV}}\\ &= \frac{6.63\times10^{-34}}{\sqrt{2\times9.11\times10^{-31}\times1.60\times10^{-19}\times64.0}}\\ &= 1.53\times10^{-10}\ \mathrm{m} = 1.53\ \mathrm{\mathring{A}}\end{aligned}

(b) Consider the interference between the atomic rows on the surface as shown in Fig. 3c.

Surface atomic rows (circles) along the [011] direction with row spacing b; two incident electron beams e- at angle phi-o to the normals n and two diffracted beams 1 and 2 at angle phi; dotted construction lines show the path difference. — Fig.3c

The path difference between electron beam 1 and 2 is

\Delta\ell = b(\sin\phi - \sin\phi_{\circ}) = n\lambda

Given $\phi_{\circ} = 15.0^{\circ}$ , $\lambda = 1.53\ \mathrm{\mathring{A}}$ and $b = \frac{a}{\sqrt{2}} = \frac{3.92}{\sqrt{2}} = 2.77\ \mathrm{\mathring{A}}$ , two solutions are possible.

(i) When $n = 0$ , $\phi = \phi_{\mathrm{o}} = 15.0^{\circ}$ (Answer 1)

(ii) When $n = 1$

\Delta\ell = 2.77(\sin\phi - \sin15^{\circ}) = 1\times1.53

\sin\phi = \frac{1.53 + 0.72}{2.77} = 0.812

\phi = 54.3^{\circ}

(Answer 2)

For $n$ = 2, no solution exists as $\Delta\ell = 2.77(sin\phi - sin15^{\circ}) = 2\times1.53$ and $sin\phi > 1$ .

Vector diagram above a horizontal line labelled surface: incident wave vector K pointing down to the surface, reflected wave vector K', and the vertical difference vector ΔK along the normal; the angle θ is marked between K and ΔK. — Fig. 3d

(2) $I = I_0\ exp\langle -(\vec{u}\cdot\Delta\vec{K})^2\rangle$

For the specularly reflected beam, we have from Fig. 3d

\Delta\vec{K} = \vec{K}' - \vec{K} = 2K\cos\theta\quad\hat{x}

where $\hat{x}$ is the unit vector in the direction of the surface normal. Take the $x$ -component of $\vec{u}$ , we then obtain

I = I_0 e^{-\lt u_x^2(t)\cdot 4K^2\cos^2\theta\gt} = I_0 e^{-4K^2\cos^2\theta\lt u_x^2(t)\gt} \qquad (2)

The vibration in the direction of the surface normal of the surface atoms is simple harmonic, take

u_x(t) = A\cos\omega t

\because\qquad \langle u_x^2(t)\rangle = \frac{1}{\tau}\int_0^{\tau} u^2\,dt = \frac{1}{\tau}\int_0^{\tau} A^2\cos^2\omega t\,dt = \frac{A^2}{\tau}\cdot\frac{\tau}{2} = \frac{A^2}{2}

\therefore\quad A^2 = 2\langle u_x^2(t)\rangle

The total energy $E$ is thus given by

E = \frac{1}{2}CA^2 = \frac{1}{2}C\cdot 2\lt u_x^2(t)\gt\ =\ C\lt u_x^2(t)\gt\ =\ m'\omega^2\lt u_x^2(t)\gt

Therefore, one obtains

\lt u_x^2(t)\gt\ =\ E/(m'\omega^2)

E = m'\omega^2\lt u_x^2\gt = k_BT

where $m'$ is the mass of the atom. From either of the above two equations, one then has the following equality

\lt u_x^2\gt = \frac{k_BT}{m'\omega^2} = \frac{k_BT}{m'4\pi^2f^2} \qquad (3)

From eq. (3) and eq. (2), one obtains

I = I_0 e^{-4K^2\cos^2\theta\frac{k_BT}{m'4\pi^2f^2}}

where $K = \frac{2\pi p}{h} = \frac{2\pi}{\lambda}$ . Accordingly,

I = I_0 e^{-\frac{4k_B\cos^2\theta}{m'f^2\lambda^2}T} = I_0 e^{-M'T} \qquad (4)

and

\ell n\frac{I}{I_0} = -M'T

From the plot of $\ell n\frac{I}{I_0}$ versus $T$ , one obtains the slope

M' = \frac{4k_B\cos^2\theta}{m'f^2\lambda^2} \qquad (5)

The slope of the curve can be estimated from Fig. 3b and leads to the result

M' = 2.3\times10^{-3}.

Using the following data in Eq.(5),

k_B = 1.38\times10^{-23}\ \mathrm{J/K}

\lambda = 1.53\times10^{-10}\ \mathrm{m}

m' = 195.1\times10^{-3}/(6.02\times10^{23}) = 3.24\times10^{-25}\ \mathrm{kg/atom}

one finds

2.3\times10^{-3} = \frac{4\times1.38\times10^{-23}\cdot cos^2 15^{\circ}}{\frac{195.1\times10^{-3}}{6.02\times10^{23}}\times f^2\times(1.53\times10^{-10})^2}

The solution for frequency is then

$f^2 = 3.0\times10^{24}$ (new) $\Rightarrow\quad f = 1.7\times10^{12}\ \mathrm{Hz}$ Answer (a)

From $\lt u_x^2\gt = \frac{k_BT}{m'4\pi^2f^2}$ , $T = 300\mathrm{K}$ , one finally obtains

$\lt u_x^2\gt = \dfrac{1.38\times10^{-23}\cdot300}{\frac{195.1\times10^{-3}}{6.02\times10^{23}}\times4\pi^2\times3.0\times10^{24}} = 1.1\times10^{-22}\ \mathrm{m}^2$ (new)

and

$\sqrt{\lt u_x^2\gt} = 1.0\times10^{-11}\ \mathrm{m} = 0.10\ \mathrm{\mathring{A}}$ (new)

Ans

Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf