# How Does Changing Lattice Dimensions Affect X-ray Scattering Angles?

• skrat
In summary: K=\frac{2\pi}{a}(m_1,m_2,\frac a b m_3)$$This is because the reciprocal lattice vectors change when the lattice constant changes, and in this case, we are changing the vertical sides of the cubic lattice from ##a## to ##b##. So your final equation for ##\Theta## should be$$\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+m_3^2}{7}})$$In summary, skrat ## Homework Statement We scatter a roentgen light with ##\lambda =\frac{2a}{\sqrt 7}## on a particle of dust of a crystal, with cubic crystal lattice as shown in the attached image. a) Calculate geometric structure factor. Calculate with cubic primitive cell. b) At which scattering angles do we find constructive interference? c) Which of these angles is changed if the vertical sides of the cubic lattice are changed from ##a## to ##b##? ## Homework Equations ## The Attempt at a Solution Ok I did something, but I am not sure if what I did is OK: a) Primitive vectors:$$\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}\\
\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}})$$and from here it should be obvious that only scattering on plane ##(1,1,1)## will change.At least I hope so.. ? Any feedback would be nice.Thanks.A:It looks like it's all correct, except for one thing.In part (c), you should include the fact that the reciprocal vector is changed to$$\vec K=\frac{2\pi}{b}(m_1,m_2,m_3)

## 1. What is "scattering" on dust particles?

"Scattering" on dust particles refers to the process of light being redirected in different directions as it interacts with dust particles in the air. This can cause the light to appear diffused or scattered, making objects appear less distinct or hazy.

## 2. What causes scattering on dust particles?

Scattering on dust particles is caused by the interaction between light and small particles of dust or other small particles suspended in the air. These particles can be natural, such as pollen or sea salt, or man-made, such as pollution or smoke.

## 3. How does scattering on dust particles affect visibility?

Scattering on dust particles can reduce visibility by making objects appear less distinct or hazy. This can be especially noticeable in areas with high levels of pollution or dust, where the scattering of light can create a visible haze in the air.

## 4. Does scattering on dust particles have any impact on health?

Inhaling dust particles can have negative impacts on health, especially for those with respiratory issues. However, the scattering of light on dust particles does not have a direct impact on health, although it can contribute to reduced visibility and air pollution.

## 5. Can scattering on dust particles be controlled or reduced?

Scattering on dust particles can be reduced by controlling the amount of dust or other particles in the air, such as through air filtration systems or reducing pollution levels. However, it is a natural phenomenon that cannot be completely eliminated.

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