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My textbook, Solid-State Physics, Fluidics, and Analytical Techniques in Micro- and Nanotechnology, by Madou, says the following in a section on X-Ray Intensity and Structure Factor $F(hkl)$:

In Figure 2.28 we have plotted $y = \dfrac{\sin^2(Mx)}{\sin^2(x)}$. This function is virtually zero except at the points where $x = n\pi$ ($n$ is an integer including zero), where it rises to the maximum value of $M^2$. The width of the peaks and the prominence of the ripples are inversely proportional to $M$.

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Remember that there are three sums in Equation 2.38. For simplicity we only evaluated one sum to calculate the intensity in Equation 2.39. The total intensity equals:

$$I \propto \dfrac{\sin^2 \left( \dfrac{1}{2} M \mathbf{a}_1 \cdot \Delta \mathbf{k} \right)}{ \sin^2 \left( \dfrac{1}{2} \mathbf{a}_1 \cdot \Delta \mathbf{k} \right)} \times \dfrac{\sin^2 \left( \dfrac{1}{2} N \mathbf{a}_2 \cdot \Delta \mathbf{k} \right)}{ \sin^2 \left( \dfrac{1}{2} \mathbf{a}_2 \cdot \Delta \mathbf{k} \right)} \times \dfrac{\sin^2 \left( \dfrac{1}{2} P \mathbf{a}_3 \cdot \Delta \mathbf{k} \right)}{ \sin^2 \left( \dfrac{1}{2} \mathbf{a}_3 \cdot \Delta \mathbf{k} \right)} \tag{2.40}$$

so that the diffracted intensity will equal zero unless all three quotients in Equation 2.40 take on their maximum values at the same time. This means that the three arguments of the sine terms in the denominators must be simultaneously equal to integer multiples of $2\pi$, or the peaks occur only when:

$$\mathbf{a}_1 \cdot \Delta \mathbf{k} = 2 \pi e$$ $$\mathbf{a}_2 \cdot \Delta \mathbf{k} = 2 \pi f$$ $$\mathbf{a}_3 \cdot \Delta \mathbf{k} = 2 \pi g$$

These are, of course, the familiar Laue equations.

I could be mistaken, but I see two possible errors here:

  1. Since we have that the function is virtually zero except at the points where $x = n \pi$, where $n$ is an integer, we use L'Hôpital's rule to get that $\dfrac{2M \cos(Mx)}{2\cos(x)} = \dfrac{M \cos(Mx)}{\cos(x)}$, which is a maximum of $M$ -- not $M^2$ -- for $x$.

  2. Assuming that we require that the arguments of the sine terms of the three denominators equal integer multiples of $2\pi$, we have that

$$\dfrac{1}{2} \mathbf{a}_1 \cdot \Delta \mathbf{k} = 2\pi e \Rightarrow \mathbf{a}_1 \cdot \Delta \mathbf{k} = 4 \pi e$$

However, as the author indicates, the Laue equation is $\mathbf{a}_1 \cdot \Delta \mathbf{k} = 2 \pi e$. So should it not be the case that we require that the arguments of the sine terms of the three denominators equal integer multiples of $\pi$, so that we have that

$$\dfrac{1}{2} \mathbf{a}_1 \cdot \Delta \mathbf{k} = \pi e \Rightarrow \mathbf{a}_1 \cdot \Delta \mathbf{k} = 2\pi e$$

I would greatly appreciate it if people would please take the time to review this.

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On applying L'Hôpital's rule, we get $ y = \frac{2Msin(Mx)cos(Mx)}{2sin(x)cos(x)}$. Again applying L'Hôpital's rule $\frac{sin(Mx)}{sin(x)} = M$, giving $y=M^{2}$.

Just here, it is proved that $\frac{sin^{2}(Mx)}{sin^{2}x}$ has maxima at $ x = n\pi$. Here n is any integer and not just even integers. In $\frac{sin^{2}(\frac{1}{2}M\vec{a_{1}}.\vec{\Delta k)}}{sin^{2}(\frac{1}{2}\vec{a_{1}}.\vec{\Delta k)}}$, $ x = \frac{1}{2}\vec{a_{1}}.\vec{\Delta k}$, therefore $$\dfrac{1}{2} \mathbf{a}_1 \cdot \Delta \mathbf{k} = \pi e \Rightarrow \mathbf{a}_1 \cdot \Delta \mathbf{k} = 2 \pi e$$

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  • $\begingroup$ Thanks for the answer. I made an error in my calculation. If we apply L'Hôpital's rule twice, then don't we get $$\dfrac{M^2[ \cos^2(Mx) - \sin^2(Mx)]}{[\cos^2(x) - \sin^2(x)]}$$? $\endgroup$ – The Pointer Jan 22 at 6:55
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    $\begingroup$ Yes this simplifies to $\frac{M^{2}cos(2Mx)}{cos(2x)}$ which in the limit is $M^{2}$ $\endgroup$ – Hari Jan 22 at 6:59
  • $\begingroup$ Ahh, yes, you're right: $$\cos(2x) = \cos^2(x) - \sin^2(x)$$ www2.clarku.edu/faculty/djoyce/trig/identities.html $\endgroup$ – The Pointer Jan 22 at 7:02
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    $\begingroup$ In the limit $x \rightarrow n\pi$ which is what we are interested in $\endgroup$ – Hari Jan 22 at 7:16
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    $\begingroup$ Remember that M is typically the number of unit cells in the lattice which is an integer, hence the cosine function is calculated at $ 2Mn\pi$ which is a multiple of $2\pi$ where the value of the cosine function is 1. $\endgroup$ – Hari Jan 22 at 7:25

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