Съдържание

IPhO 1990, Теория и експеримент — Задача 1. X-ray Diffraction from a crystal.

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Всички материали от IPhO 1990

THE 21st INTERNATIONAL PHYSICS OLYMPIAD - 1990 · 10 т.

Внимание: Бележка към темата

Complete combined archive booklet, attributed to Hans Jordens, contains three theory questions and two experiments with keys. Printed errors and gaps are preserved: the neutron-star key changes the flattening denominator and uses a square-law radius inference; the LED key references missing graphs, has printed inconsistent table values and one unassigned final value; magnet-speed tables print m/s but their values match cm/s. Native source values and all grading remain unchanged.

Условие

We wish to study X-ray diffraction by a cubic crystal lattice. To do this we start with the diffraction of a plane, monochromatic wave that falls perpendicularly on a 2-dimensional grid that consists of $N_1\times N_2$ slits with separations $d_1$ and $d_2$ . The diffraction pattern is observed on a screen at a distance $L$ from the grid. The screen is parallel to the grid and $L$ is much larger than $d_1$ and $d_2$ .

a- Determine the positions and widths of the principal maximum on the screen.
The width is defined as the distance between the minima on either side of the maxima. [4 т.]

We consider now a cubic crystal, with lattice spacing $a$ and size $N_0a\times N_0a\times N_1a$ . $N_1$ is much smaller than $N_0$ . The crystal is placed in a parallel X-ray beam along the z-axis at an angle $\Theta$ (see Fig. 1). The diffraction pattern is again observed on a screen at a great distance from the crystal.

Cubic crystal, axes x y z, incident X-ray beam, angle Θ and dimensions N0a and N1a. — Figure 1. Diffraction of a parallel X-ray beam along the z-axis. The angle between the crystal and the y-axis is $\Theta$ .

b- Calculate the position and width of the maxima as a function of the angle $\Theta$ (for small $\Theta$ ).

What in particular are the consequences of the fact that $N_1\ll N_0$ . [3 т.]

The diffraction pattern can also be derived by means of Bragg's theory, in which it is assumed that the X-rays are reflected from atomic planes in the lattice. The diffraction pattern then arises from interference of these reflected rays with each other.

c- Show that this so-called Bragg reflection yields the same conditions for the maxima as those that you found in b. [2 т.]

In some measurements the so-called powder method is employed. A beam of X-rays is scattered by a powder of very many, small crystals. (Of course the sizes of the crystals are much larger than the lattice spacing, $a$ ).
Scattering of X-rays of wavelength $0.15\,\mathrm{nm}$ by Potassium Chloride [KCl] (which has a cubic lattice, see Fig.2) results in the production of concentric dark circles on a photographic plate. The distance between the crystals and the plate is $0.10\,\mathrm m$ , and the radius of the smallest circle is $0.053\,\mathrm m$ (see Fig. 3). $\mathrm{K^+}$ and $\mathrm{Cl^-}$ ions have almost the same size, and they may be treated as identical scattering centres.

Native KCl cubic lattice with light and dark ions. — Figure 2. The cubic latice of Potassium Chloride in which the $\mathrm{K^+}$ and $\mathrm{Cl^-}$ wons have almost the same size.

Concentric diffraction circles on the rectangular photographic plate; radius arrow. — Figure 3. Scattering of X-rays by a powder of KCl crystals results in the production of concentric dark circles on a photographic plate.

d- Calculate the distance between two neighbouring K ions in the crystal. [1 т.]

Решение

a-

Consider first the x-direction. If waves coming from neighbouring slits (with separation $d_1$ ) traverse paths of lengths that differ by:

\Delta_1=n_1\lambda

where $n_1$ is an integer, then a principal maximum occurs. The position on the screen (in the x-direction) is:

x_{n_1}=\frac{n_1\lambda L}{d_1}

since $d_1\ll d_2$ .
The path difference between the middle slit and one of the slits at the edge is then:

\Delta_{(N_1/2)}=\frac{N_1}{2}n_1\lambda

If on the other hand this path difference is:

\Delta_{(N_1/2)}=\frac{N_1}{2}n_1\lambda+\frac\lambda2

then the first minimum, next to the principal maximum, occurs. The position of this minimum on the screen is given by:

x_{n_1}+\Delta x=\frac{(\frac{N_1}{2}n_1\lambda+\frac\lambda2)L}{\frac{N_1}{2}d_1}=\frac{n_1\lambda L}{d_1}+\frac{\lambda L}{N_1d_1}

\rightarrow\quad\Delta x=\frac{\lambda L}{N_1d_1}

The width of the principal maximum is accordingly:

2\Delta x=2\frac{\lambda L}{N_1d_1}

A similar treatment can be made for the y-direction, in which there are $N_2$ slits with separation $d_2$ . The positions and widths of the principal maximal are:

(x_{n_1},y_{n_2})=\left(\frac{n_1\lambda L}{d_1},\frac{n_2\lambda L}{d_2}\right)

2\Delta x=2\frac{\lambda L}{N_1d_1};\qquad2\Delta y=2\frac{\lambda L}{N_2d_2}

An alternative method of solution is to calculate the intensity for the 2-dimensional grid as a function of the angle that the beam makes with the screen.

b-

In the x-direction the beam 'sees' a grid with spacing $a$ , so that in this direction we have:

x_{n_1}=\frac{n_1\lambda L}{a}\qquad\Delta x=2\frac{\lambda L}{N_0a}

In the y-direction, the beam 'sees' a grid with effective spacing $a\cos(\Theta)$ . Analogously, we obtain:

y_{n_2}=\frac{n_2\lambda L}{a\cos(\theta)}\qquad\Delta y=2\frac{\lambda L}{N_0a\cos(\theta)}

In the z-direction, the beam 'sees' a grid with effective spacing $a\sin(\Theta)$ . This gives rise to principal maxima with position and width:

y'_{n_3}=\frac{n_3\lambda L}{a\sin(\theta)}\qquad\Delta y'=2\frac{\lambda L}{N_1a\sin(\theta)}

This pattern is superimposed on the previous one. Since $\sin(\Theta)$ is very small, only the zeroth-order pattern will be seen, and it is very broad, since $N_1\sin(\Theta)\ll N_0$ . The diffraction pattern from a plane wave falling on a thin plate of a cubic crystal, at a small angle of incidence to the normal, will be almost identical to that from a two-dimensional grid.

c-

In Bragg reflection, the path difference for constructive interference between neighbouring planes:

\Delta=2a\sin(\phi)\approx2a\phi=n\lambda\quad\rightarrow\quad\frac{x}{L}\approx2\phi\approx\frac{n\lambda}{a}\quad\rightarrow\quad x\approx\frac{n\lambda L}{a}

Here $\phi$ is the angle of diffraction.
This is the same condition for a maximum as in section b.

d-

For the distance, $\sqrt2a$ , between neighbouring K ions we have:

\operatorname{tg}(2\phi)=\frac{x}{L}=\frac{0{,}053}{0{,}1}\approx0{,}53\quad\rightarrow\quad a=\frac\lambda{2\sin(\phi)}\approx\frac{0{,}15\cdot10^{-9}}{2\cdot0{,}24}\approx0{,}31\,\mathrm{nm}

K-K\approx\sqrt2\cdot0{,}31\approx0{,}44\,\mathrm{nm}

Marking Breakdown

Part	Criterion	Marks
a	position of principal maxima	1
a	width of principal maxima	3
b	lattice constants	1
b	effect of thickness	2
c	Bragg reflection	2
d	Calculation of K-K spacing	1

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