IPhO 1990 · Theory problem 1

X-ray Diffraction from a crystal

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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.

We consider now a cubic crystal, with lattice spacing \(a\) and size \(N_0.a \times N_0.a \times N_1.a\). \(N_1\) 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 great distance from the crystal.

Figure 1

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 \(N_1 \ll N_0\).

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.

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.

Figure 2

Figure 2. The cubic lattice of Potassium Chloride in which the \(\mathrm{K}^{+}\) and \(\mathrm{Cl}^{-}\) ions have almost the same size.

Figure 3

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.

Official English text of the problem.

Answers

Let \(\lambda\) denote the wavelength. Screen coordinates \(X,Y\) are measured from the undeflected beam; for the crystal, \(L\) denotes the crystal-to-screen distance.

  • a. With integers \(m,n\), put \[ u_m=\frac{m\lambda}{d_1},\qquad v_n=\frac{n\lambda}{d_2}. \] The principal maxima are at \[ \boxed{(X_{mn},Y_{mn}) =\frac{L(u_m,v_n)}{\sqrt{1-u_m^2-v_n^2}}.} \] Their full first-zero widths in direction cosines are \[ \boxed{\Delta u=\frac{2\lambda}{N_1d_1},\qquad \Delta v=\frac{2\lambda}{N_2d_2}.} \] For small diffraction angles, these become \[ \boxed{X_m\simeq\frac{m\lambda L}{d_1},\quad Y_n\simeq\frac{n\lambda L}{d_2},\quad W_X\simeq\frac{2\lambda L}{N_1d_1},\quad W_Y\simeq\frac{2\lambda L}{N_2d_2}.} \]

  • b. Write \(r=\lambda/a\), and let \(h,k,\ell\) be integer diffraction indices. Define \[ B=k\sin\Theta+\ell\cos\Theta,\qquad C=k\cos\Theta-\ell\sin\Theta. \] Fully constructive maxima satisfy \[ \boxed{2B+r(h^2+k^2+\ell^2)=0} \] and appear at \[ \boxed{(X,Y)=\frac{Lr(h,C)}{1+rB}.} \] For small \(\Theta\), use \(B\simeq k\Theta+\ell\) and \(C\simeq k-\ell\Theta\).

For forward diffraction at small scattering angles, the sharp spots have \[ \boxed{(X,Y)\simeq Lr(h,k),\qquad W_X\simeq W_Y\simeq\frac{2\lambda L}{N_0a}.} \] Their thickness interference factor is maximal when \[ \boxed{-k\Theta-\frac{\lambda}{2a}(h^2+k^2)=\ell,} \] with adjacent zeros when the left-hand side equals \(\ell\pm1/N_1\). Thus \(N_1\ll N_0\) gives narrow spots with a much broader condition for their excitation. For \(k\ne0\), the corresponding full width in tilt is approximately \[ \boxed{\Delta\Theta=\frac{2}{N_1|k|}.} \]

  • c. Bragg’s condition \[ \boxed{2d\sin\theta_B=p\lambda} \] is equivalent to the lattice-interference conditions in b. In particular, reflection from planes normal to the tilted long crystal axis gives \[ \boxed{2a\sin\Theta=p\lambda,\qquad X=0,\quad Y=-L\tan(2\Theta)} \] for the orientation chosen below.

  • d. The distance between nearest neighbouring potassium ions is \[ \boxed{d_{\mathrm{K-K}}=4.4\times10^{-10}\ \mathrm{m} =0.44\ \mathrm{nm}.} \]

Solution

We use coherent, elastic scattering in the far-field—or Fraunhofer—approximation. The slits and lattice sites are treated as identical radiators; their individual diffraction envelope is omitted.

A useful result for \(N\) equally spaced radiators with phase difference \(2\pi t\) is \[ F_N(t)=\left|\sum_{j=0}^{N-1}e^{2\pi ijt}\right|^2 =\frac{\sin^2(N\pi t)}{\sin^2(\pi t)}. \] At an integer \(t=m\), the limiting value is \(N^2\): all contributions are in phase. For \(N>1\), the nearest zeros are \[ t=m\pm\frac1N. \]

a. Diffraction by the two-dimensional grid

Take the grid axes along \(x,y\), with the incident wave along \(z\). An outgoing ray has unit direction \[ \mathbf s=(u,v,w),\qquad w=\sqrt{1-u^2-v^2}>0. \]

The incident phase is the same at every slit. The outgoing path differences between adjacent slits are \(d_1u\) and \(d_2v\). Hence \[ I(u,v)\propto F_{N_1}\!\left(\frac{d_1u}{\lambda}\right) F_{N_2}\!\left(\frac{d_2v}{\lambda}\right). \]

Both factors attain their principal maxima when \[ d_1u=m\lambda,\qquad d_2v=n\lambda. \] Only orders satisfying \(u_m^2+v_n^2<1\) reach a screen in the forward half-space.

A ray in direction \(\mathbf s\) intersects the plane \(z=L\) at \[ X=L\frac{u}{w},\qquad Y=L\frac{v}{w}. \] Therefore \[ (X_{mn},Y_{mn}) =\frac{L(u_m,v_n)}{\sqrt{1-u_m^2-v_n^2}}. \]

The nearest zero boundaries are \[ u=u_m\pm\frac{\lambda}{N_1d_1}, \qquad v=v_n\pm\frac{\lambda}{N_2d_2}. \] These equations, together with the screen mapping above, specify the first-zero boundaries without a small-angle approximation.

For example, define \[ \mathbf R(u,v)=\frac{L(u,v)}{\sqrt{1-u^2-v^2}}. \] The two pairs of first-zero points along the natural grid coordinates are \[ \mathbf R\!\left(u_m\pm\frac{\lambda}{N_1d_1},v_n\right), \qquad \mathbf R\!\left(u_m,v_n\pm\frac{\lambda}{N_2d_2}\right). \] Their respective separations give the corresponding screen widths.

For small diffraction angles, \(u,v\ll1\), so \(X\simeq Lu\), \(Y\simeq Lv\). Thus \[ X_m\simeq\frac{m\lambda L}{d_1},\qquad Y_n\simeq\frac{n\lambda L}{d_2}, \] and \[ \boxed{W_X\simeq\frac{2\lambda L}{N_1d_1},\qquad W_Y\simeq\frac{2\lambda L}{N_2d_2}.} \]

b. Diffraction by the tilted finite crystal

Choose crystal unit vectors \[ \mathbf e_1=\hat{\mathbf x},\qquad \mathbf e_2=\hat{\mathbf y}\cos\Theta+\hat{\mathbf z}\sin\Theta, \qquad \mathbf e_3=-\hat{\mathbf y}\sin\Theta+\hat{\mathbf z}\cos\Theta. \] The two long dimensions are along \(\mathbf e_1,\mathbf e_2\); the thin dimension is along \(\mathbf e_3\). This takes positive \(\Theta\) as a tilt of the long crystal axis from \(+y\) towards \(+z\).

The incident and outgoing wavevectors are \[ \mathbf k_i=\frac{2\pi}{\lambda}\hat{\mathbf z},\qquad \mathbf k_f=\frac{2\pi}{\lambda}\mathbf s. \] Define the scattering vector \[ \mathbf q=\mathbf k_f-\mathbf k_i. \]

Unlike the grid in a, atoms at different depths also have different incident phases. Both incident and outgoing propagation are accounted for by \(\mathbf q\). Define \[ p_i=\frac{a}{2\pi}\mathbf q\cdot\mathbf e_i. \] Explicitly, \[ \begin{aligned} p_1&=\frac{a}{\lambda}s_x,\\ p_2&=\frac{a}{\lambda} \left[s_y\cos\Theta+(s_z-1)\sin\Theta\right],\\ p_3&=\frac{a}{\lambda} \left[-s_y\sin\Theta+(s_z-1)\cos\Theta\right]. \end{aligned} \] The complete finite-lattice interference pattern is therefore \[ \boxed{ I(\mathbf s)\propto F_{N_0}(p_1)\,F_{N_0}(p_2)\,F_{N_1}(p_3). } \]

A fully constructive maximum requires \[ p_1=h,\qquad p_2=k,\qquad p_3=\ell, \] where \(h,k,\ell\) are integers. Equivalently, \[ \mathbf q=\frac{2\pi}{a} (h\mathbf e_1+k\mathbf e_2+\ell\mathbf e_3). \] Writing \(r=\lambda/a\), the outgoing direction is \[ \begin{aligned} s_x&=rh,\\ s_y&=r(k\cos\Theta-\ell\sin\Theta),\\ s_z&=1+r(k\sin\Theta+\ell\cos\Theta). \end{aligned} \]

Elastic scattering requires \(|\mathbf s|=1\). Substitution gives \[ \boxed{ 2(k\sin\Theta+\ell\cos\Theta) +r(h^2+k^2+\ell^2)=0. } \] For any allowed forward order, its screen position is \[ \boxed{ X=\frac{Lrh}{1+r(k\sin\Theta+\ell\cos\Theta)},\qquad Y=\frac{Lr(k\cos\Theta-\ell\sin\Theta)} {1+r(k\sin\Theta+\ell\cos\Theta)}. } \]

These formulas retain the exact outgoing-ray geometry. For small crystal tilt, \[ 2(k\Theta+\ell)+r(h^2+k^2+\ell^2)\simeq0, \] and \[ X\simeq\frac{Lrh}{1+r(\ell+k\Theta)},\qquad Y\simeq\frac{Lr(k-\ell\Theta)}{1+r(\ell+k\Theta)}. \]

The first zeros surrounding a fully constructive maximum occur on \[ p_1=h\pm\frac1{N_0},\qquad p_2=k\pm\frac1{N_0},\qquad p_3=\ell\pm\frac1{N_1}. \] Thus the full widths in scattering-vector components are \[ \boxed{ \Delta q_1=\Delta q_2=\frac{4\pi}{N_0a}, \qquad \Delta q_3=\frac{4\pi}{N_1a}. } \]

Because \(N_1\ll N_0\), the first two factors form narrow peaks, while the third factor supplies a relatively broad envelope. Consequently, a sharp spot can remain visible even when \(p_3\) is not exactly an integer.

For completeness, the position of such a spot can be obtained without requiring exact three-dimensional constructive interference. Set \(p_1=h,\ p_2=k\). On the forward branch, \[ \zeta_{hk}(\Theta)= \frac{ \sqrt{1-r^2h^2-(\sin\Theta+rk)^2}-\cos\Theta }{r} \] is the resulting value of \(p_3\). Its position is \[ \boldsymbol{\mathcal R}_\Theta(h,k) = \frac{Lr\bigl(h,\ k\cos\Theta-\zeta_{hk}\sin\Theta\bigr)} {1+r(k\sin\Theta+\zeta_{hk}\cos\Theta)}. \] Its intensity contains the factor \(F_{N_1}(\zeta_{hk})\). Fully constructive peaks occur when \(\zeta_{hk}\) is an integer.

For a well-excited spot, the two pairs of first-zero points are obtained from this mapping by \[ h\longrightarrow h\pm\frac1{N_0}, \qquad\text{or}\qquad k\longrightarrow k\pm\frac1{N_0}. \] Their separations give the screen widths, including the dependence on \(\Theta\).

The small-angle form makes the physical consequences especially clear. Put \[ \alpha=\frac XL,\qquad \beta=\frac YL, \qquad \mathbf s\simeq \left(\alpha,\beta,1-\frac{\alpha^2+\beta^2}{2}\right). \] For small \(\Theta,\alpha,\beta\), \[ p_1\simeq\frac{\alpha}{r},\qquad p_2\simeq\frac{\beta}{r},\qquad p_3\simeq-\frac{\Theta\beta+(\alpha^2+\beta^2)/2}{r}. \] The quadratic angular term must be retained in \(p_3\), since it is the leading longitudinal path difference.

The narrow transverse peaks therefore occur at \[ \alpha_h\simeq hr,\qquad \beta_k\simeq kr, \] with full widths \[ \Delta\alpha\simeq\Delta\beta\simeq\frac{2r}{N_0}. \] Hence \[ \boxed{ (X,Y)\simeq\left(\frac{h\lambda L}{a},\frac{k\lambda L}{a}\right), \qquad W_X\simeq W_Y\simeq\frac{2\lambda L}{N_0a}. } \]

At these positions, \[ p_3\simeq-k\Theta-\frac r2(h^2+k^2). \] Thus a thickness-envelope maximum occurs when \[ -k\Theta-\frac r2(h^2+k^2)=\ell, \] and its adjacent zeros occur at \(\ell\pm1/N_1\).

For the central thickness order \(\ell=0\), the strongest spots satisfy \[ h^2+\left(k+\frac{\Theta}{r}\right)^2 =\left(\frac{\Theta}{r}\right)^2. \] Finite thickness broadens this circle into a band of appreciably excited lattice spots.

For \(k\ne0\), varying the tilt changes \(p_3\) at the rate \[ \frac{dp_3}{d\Theta}\simeq-k. \] The full first-zero width in tilt is therefore \[ \boxed{\Delta\Theta\simeq\frac{2}{N_1|k|}.} \]

The two large dimensions determine the narrow spot widths. The thin dimension gives a broader interference envelope and a broader range of crystal orientations over which a spot is appreciably excited.

c. Equivalence with Bragg reflection

Consider a family of atomic planes with coprime integer indices \((H,K,M)\). Their unit normal and spacing are \[ \hat{\mathbf n} =\frac{H\mathbf e_1+K\mathbf e_2+M\mathbf e_3} {\sqrt{H^2+K^2+M^2}}, \qquad d=\frac{a}{\sqrt{H^2+K^2+M^2}}. \]

Let \(\theta_B\) be the glancing angle between the incident beam and these planes. Rays reflected from consecutive planes have path difference \[ 2d\sin\theta_B. \] Constructive interference requires \[ \boxed{2d\sin\theta_B=p\lambda,\qquad p=1,2,\ldots.} \]

Specular reflection changes only the component of the wavevector normal to the planes. Therefore \(\mathbf q\) is normal to them and has magnitude \[ |\mathbf q|=2\frac{2\pi}{\lambda}\sin\theta_B. \] Using Bragg’s condition, \[ |\mathbf q|=\frac{2\pi p}{d}. \] With a suitable choice of normal direction, \[ \mathbf q =\frac{2\pi p}{a} (H\mathbf e_1+K\mathbf e_2+M\mathbf e_3). \] Setting \[ (h,k,\ell)=p(H,K,M) \] gives exactly \[ \mathbf q=\frac{2\pi}{a} (h\mathbf e_1+k\mathbf e_2+\ell\mathbf e_3), \] the three lattice-interference conditions derived in b. Enforcing equal incident and outgoing wavelengths then gives the same condition \[ 2(k\sin\Theta+\ell\cos\Theta) +\frac{\lambda}{a}(h^2+k^2+\ell^2)=0. \]

As a direct example, take planes normal to \(\mathbf e_2\). Their spacing is \(a\), and their glancing angle is \(\Theta\). Bragg’s law gives \[ 2a\sin\Theta=p\lambda. \] In b, choose \(h=0,\ k=-p,\ \ell=0\). The condition becomes \[ -2p\sin\Theta+\frac{\lambda}{a}p^2=0, \] which is identical. The outgoing ray is \[ \mathbf s=(0,-\sin2\Theta,\cos2\Theta), \] so \[ X=0,\qquad Y=-L\tan2\Theta. \]

d. Nearest-neighbour potassium-ion distance

When \(\mathrm K^+\) and \(\mathrm{Cl}^-\) scatter identically, the alternating ionic arrangement acts as a simple cubic lattice of identical scattering centres.

Let \(b\) denote the nearest-neighbour K–Cl separation. The effective simple cubic lattice spacing is \(b\). Its largest reflecting-plane spacing is \(b\), so the smallest nonzero diffraction angle comes from first-order reflection with \[ 2b\sin\theta_B=\lambda. \]

Random crystal orientations produce a cone of scattered rays with scattering angle \(2\theta_B\). A perpendicular photographic plate therefore records a ring of radius \[ R=L\tan(2\theta_B). \] Using the given data, \[ 2\theta_B=\arctan\left(\frac{0.053}{0.10}\right) =27.9236^\circ, \] and hence \[ \theta_B=13.9618^\circ. \] No small-angle approximation is needed here.

Thus \[ b=\frac{0.15\ \mathrm{nm}} {2\sin13.9618^\circ} =0.31085\ \mathrm{nm}. \]

The species alternate along each simple cubic axis. Two nearest potassium ions therefore lie across a face diagonal of a cube of side \(b\), giving \[ d_{\mathrm{K-K}}=\sqrt2\,b =\sqrt2(0.31085\ \mathrm{nm}) =0.43961\ \mathrm{nm}. \] To two significant figures, \[ \boxed{d_{\mathrm{K-K}}=0.44\ \mathrm{nm} =4.4\times10^{-10}\ \mathrm{m}.} \]

The solution was checked against the official solution.