# Homework Help: Scattering on dust particle

1. Apr 12, 2015

### skrat

1. The problem statement, all variables and given/known data
We scatter a roentgen light with $\lambda =\frac{2a}{\sqrt 7}$ on a particle of dust of a crystal, with cubic crystal lattice as shown in the attached image.

a) Calculate geometric structure factor. Calculate with cubic primitive cell.
b) At which scattering angles do we find constructive interference?
c) Which of these angles is changed if the vertical sides of the cubic lattice are changed from $a$ to $b$?

2. Relevant equations

3. The attempt at a solution
Ok I did something, but I am not sure if what I did is OK:

a)
Primitive vectors: $$\vec a_1=a(1,0,0)$$ $$\vec a_2=a(0,1,0)$$ $$\vec a_3=a(0,0,1)$$ and basis vectors $$\vec r_0=(0,0,0)$$ $$\vec r_1=\frac a 2(1,1,0)$$ Structure factor is defined as $$S=\sum _n e^{i\vec K\vec r_n}$$ where reciprocal vector is $\vec K =\frac{2\pi}{a}(m_1,m_2,m_3)$, where I already used Miller indexes.
Finally $$S=1+e^{i\vec K \vec r_1}=1+e^{i\pi (m_1+m_2)}= \left\{\begin{matrix} 2 ,& \qquad m_1\text{ and }m_2 \text{ both odd or even}\\ 0 ,& \qquad \text{otherwise} \end{matrix}\right.$$

b) According to Bragg's Law: $$\lambda =2d \sin \frac \Theta 2$$ where $d$ is the distance between the lattice planes defined as $$d=\frac{2\pi}{|\vec K|}=\frac{a}{\sqrt{m_1^2+m_2^2+m_3^2}}.$$ Because $$\sin \frac \Theta 2=\frac{\lambda}{2d}\leq 1$$ than taking the last inequation also $$\sqrt{m_1^2+m_2^2+m_3^2}\leq \frac{2a}{\lambda}=\sqrt 7$$ where I already inserted $\lambda$.
This now leaves me with only a few (at least I couldn't think of any more that would obey condition form a) and the last one written above) possible options of Miller index combination: $$(1,1,1)\rightarrow \sqrt{m_1^2+m_2^2+m_3^2}=\sqrt 3 \leq 7$$ and $$(2,0,0)\rightarrow \sqrt{m_1^2+m_2^2+m_3^2}=\sqrt 4 \leq 7 .$$ Knowing this, brings me to $$\sin \frac \Theta 2=\frac{\lambda}{2d}=\frac{\lambda }{2a}\sqrt{ m_1^2+m_2^2+m_3^2}$$ or what I really need is $$\Theta=2\arcsin (\sqrt{\frac{m_1^2+m_2^2+m_3^2}{7}}).$$ This finally brings me to $$(1,1,1)\rightarrow \Theta =81.78°$$ and $$(2,0,0)\rightarrow \Theta =98.22°.$$

c) I think this part can be quite easily done. Reciprocal vector changes $$\vec K=\frac{2\pi}{a}(m_1,m_2,m_3) \rightarrow \vec K=\frac{2\pi}{a}(m_1,m_2,\frac a b m_3).$$ If I am not mistaken one can easily in a couple of seconds find out that the only thing that changes in end result is $$\Theta=2\arcsin (\sqrt{\frac{m_1^2+m_2^2+m_3^2}{7}})\rightarrow \Theta=2\arcsin (\sqrt{\frac{m_1^2+m_2^2+(\frac a b m_3)^2}{7}})$$ and from here it should be obvious that only scattering on plane $(1,1,1)$ will change.

At least I hope so.. ?

2. Apr 17, 2015