Understanding Ewald's sphere in the context of X-Ray diffraction

##\require{physics}##I am trying to understand how the Ewald's sphere works in the context of X-ray diffraction (XRD). I am reading from Kittel's book, as well as a few lecture series. Let me first state what I have learnt in this context (please correct me if I am wrong).

• For any real lattice with translation vector ##\va{R} = n_1 \va{a}_1 + n_2 \va{a}_2 + n_3 \va{a}_3##, we can have a vector in the reciprocal space, defined as

$$\begin{align}
\mathrm{e}^{i \va{G} \vdot \va{R}} &= 1 \\
\implies \va{G} \vdot \va{R} &= 2\pi m;\ \ m \in \mathbb{Z}.
\end{align}$$ where ##\va{a}_i## are the basis vectors in the real space.

• ##\va{G}_{hk\ell} = h \va{g}_1 + k \va{g}_2 + \ell \va{g}_3## is a vector perpendicular to the set of planes in the real space whose Miller indices are given by ##(h, k, \ell);## ##\va{g}_i## are the basis vectors in the reciprocal space.

$$\begin{equation}
\abs{\va{G}_{hk\ell}} = \dfrac{2\pi}{d_{hk\ell}}
\end{equation}$$ where ##d_{hk\ell}## is the perpendicular distance between to planes with the same Miller indices ##(h, k, \ell).##

• If ##\va{k}## is the wave vector of the incident X-ray beam, and ##\va{k'}## is the wave vector of the diffracted X-ray, then the scattering vector is ##\Delta \va{k} = \va{k'} - \va{k}.##

• The following are equivalent ways of writing Bragg's Law:

$$\begin{align}
2 d_{hk\ell} \sin \theta &= n \lambda;\\
\Delta \va{k} &= \va{G}.
\end{align}$$
With this, let's come to the Ewald's sphere. I will go ahead and copy another diagram from Kittel:

The caption reads:

The points on the RHS are reciprocal lattice points of the crystal. The vector ##\va{k}## is drawn in the direction of the incident X-ray beam, and the origin is chosen such that ##\va{k}## terminates on any reciprocal lattice point (say, A). We draw a sphere of radius $$\abs{\va{k}} = \dfrac{2\pi}{\lambda}$$ about the origin of ##\va{k}##. A diffracted beam will be formed if the sphere intersects any other point in the reciprocal lattice. The sphere, as drawn, intercepts a point (say, B) connected with the end of ##\va{k}## by a reciprocal lattice vector ##\va{G}##. The diffracted beam is in the direction $$\va{k'} = \va{k} + \va{G}.$$ The angle ##\theta## is the Bragg angle of diffraction.

So, the reciprocal lattice points are basically ##(h, k, \ell).##

These are my questions:

1. The incident wave vector ##\va{k}## terminates on the point ##A(h_1, k_1, \ell_1).## In the real space, is this the set of lattice planes from where the reflection is taking place?
2. If the above is correct, what is the significance of the point ##B(h_2, k_2, \ell_2)## in the real space? This is a different set of planes. What role do these planes play in the diffraction? I understand from geometry that this construction is basically satisfying Bragg's Law in one of the forms stated above. But I couldn't understand the role of the set of planes ##(h_2, k_2, \ell_2)## in the real space.
3. Is it possible that the sphere passes through more than two points? If yes, how can we interpret this situation? For instance, if the sphere passes through a third point ##C(h_3, k_3, \ell_3),## we can connect ##A## and ##C## with a different ##\va{G}## vector. And so, there will be reflections in different directions...?

The procedure is like this - let the point "A" in the diagram denote the origin (000) in reciprocal space. Draw the incoming k-vector with its tip on the point A, as in the diagram. Then draw the sphere of radius |k| around the tail of this vector. If the sphere intersects any point B = (hkl) in reciprocal space, then the (hkl) planes satisfy the Bragg equation and you get a diffraction peak (reflection) from this plane - you can have several such planes (see e.g. x-ray powder diffraction spectra).

There's a few ways of seeing why this works, either trigonometrically: radius of sphere = 1/##\lambda##, distance between AB = ##1/d_{\mathrm{hkl}}## = 1/(hkl plane spacing), use triangle formulae,

or by realising that i) the fact the scattering is elastic implies that |k| = |k'| = 1/##\lambda##, hence the sphere construction and ii) if the point B lies on the sphere, then k-k' is a reciprocal lattice vector, which is the equivalent statement of Bragg's law.

ergospherical said:

The procedure is like this - let the point "A" in the diagram denote the origin (000) in reciprocal space. Draw the incoming k-vector with its tip on the point A, as in the diagram. Then draw the sphere of radius |k| around the tail of this vector. If the sphere intersects any point B = (hkl) in reciprocal space, then the (hkl) planes satisfy the Bragg equation and you get a diffraction peak (reflection) from this plane - you can have several such planes (see e.g. x-ray powder diffraction spectra).

There's a few ways of seeing why this works, either trigonometrically: radius of sphere = 1/##\lambda##, distance between AB = ##1/d_{\mathrm{hkl}}## = 1/(hkl plane spacing), use triangle formulae,

or by realising that i) the fact the scattering is elastic implies that |k| = |k'| = 1/##\lambda##, hence the sphere construction and ii) if the point B lies on the sphere, then k-k' is a reciprocal lattice vector, which is the equivalent statement of Bragg's law.

That answers all my questions. Thanks! Kittel should have explicitly written somewhere that point A is to be taken as the origin.

Wrichik Basu said:

Thanks! Kittel should have explicitly written somewhere that point A is to be taken as the origin.

The 'A,B' diagram is of reciprocal space; there is no origin in it (afaiu) because the origin would be at (∞,∞), so everything is relative to point A.
It's years since I did Xray diffraction and I remember having similar problems because I didn't really 'get' reciprocal space at the time.
Does that.help - or is it bloomin' obvious?