One way to get the Scherrer formula for size broadening

Reinhold Kainhofer, reinhold@kainhofer.com

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General stuff / Introduction

We look at the scattering process as scattering at a lattice where every point at the 2-dimensional lattice emits a spherical wave. All the spherical waves interfere and at most of the angles, destructive intereference will occur.

However, a peak width of 0 can only be expected for an infinite extension (meaning infinitely many interfering spherical waves). This is clear as a delta function is not a continuous function and can thus not be created by adding up finitely many continuous functions (spherical waves in this example).
For finitely many scatterers, we have to sum up all the spherical waves, and then find the angle where the intensity is half maximum. is then the FWHM.   is here the angle between the incoming wave and the vertical net (perpendicular to the lattice plane!!! This is different from what we did in class. I used this way, not because I copied this from somewhere, but because I didn't remember the formula correctly. This whole derivation of the Scherrer formula is my own work.)

As we will detect the peaks quite far away from the scattering centers, we can assume that only the parallel parts coming from each scattering center will interfere (in other words, the wave emitted from the first center with angle θ will only interefere with the wave emmited from the second center with the same angle,  or again in other words, since the rays are asymptotically parallel, we only look at this asymptotic case.).
Note that is suffices to look only at a string of vertical scatterers, because only the difference in the light's path is important, so it does not matter, where in the lattice plane the light ray is scattered.

Path difference for waves scattered at different lattice points

First, let's look at two scattered, interfering waves, and find the difference, if they are reflected with an angle of . This difference is

Since will be a very small angle, and :

Since (Bragg Equation!!!), and we will only look at relative phases, we can neglect the two terms:

What remains is to calculate the difference in the wave's ways that has half the intensity  of the constructive interference.

Path difference for half maximum

For this, we add NN (the number of scatterers, here ) waves with a relative phase of (we only look at direction , so it suffices to look at 1-dimensional waves):

Now let's calculate the sums. For convenience's sake, from now on, I substitute k*r by r (and k*r1 by r1). In the end, I will replace r by r*k again.

fs[x_,NN_]:=Sum[Sin[n x], {n,0,NN}]

fc[x_,NN_]:=Sum[Cos[n x], {n,0,NN}]

exprc=fc[r1,NN]

exprs=fs[r1,NN]

Now taking into account that a Sin[α]+b Cos[α] can be written as , this the wave into θ direction changes to:

As we look at the Amplitude, we don't need the :

As NN grows, the full width at half maximum gets smaller of course:

Plot[Evaluate[exx[r1,#]&/@{1,2,3,4,5,10,20}], {r1,0,π/2}, PlotRange->All]

-- Graphics --

To find the r1 value where half maximum occurs, I expand this expression around (for high N the half maximum will eventually lie inside the convergence radius of the seies, so I don't worry about convergence)

Approximation to different orders.

Take the higher order approximation. As the plot shows, it approximates the r quite well.

-- Graphics --

sol2=Last[sol2nd]sol4=sol4th[[2]]//Together

Finally putting everything together

Our NN are usually quite large, so only look at the highest power in sol2 and sol4:

These are the solutions for , so if we enter this in our expression above, we get:

Let's finally look at for the 6th order approximation and call this K:

0.8882787514833481`

As we used ideal materials with ideal physics and ideal scattering, in reality the peak will be more smeared out and our calculated width determined by the equation above will be a lower bound for the actual width, thus ...
However, the upper bound still remains a mystery.

Converted by Mathematica      October 29, 2000
Styles converted by a program by Reinhold Kainhofer ( reinhold@kainhofer.com )