X-ray Diffraction, Intensity vs. Wavelength Graph

In summary: K-alpha and K-beta are two types of energy that are emitted when an electron is knocked out of an atom.
  • #1
Mxbn0
3
0

Homework Statement



[PLAIN]http://b.imagehost.org/0607/Question_7.png

Homework Equations



[PLAIN]http://d.imagehost.org/0813/Untitled_6.jpg

The Attempt at a Solution



I was able to get question b, which ends up being 73pm, but as for questions a and c, I was unable to come up with an explanation, and even when I sort of did, I couldn't explain it well. For question a, I can't figure out why the carbon is there, as well as the crystal and for question c, is it to do with the x-rays hitting either 1. the electron cloud or 2. the nucleus itself (compton collision) ?
 
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  • #2
Hi,
please tell us how did you calculate 73 pm ?
 
  • #3
You find delta-lambda, equals to 2, where theta in this case is equal to 90, then you add it onto the initial value of 73pm given.
 
  • #4
Hi,
I have some doubts:
In figure i notice only theta angle, but no psi angle! [is your question taken from book or some other sources?]
for a) I guess, one can apply the famous law that relates incident angle and scattering plane's angle.
b) use the same law from (a)
c) it is a type of scattering (incident and scattered X-ray have same wavelength!)
PS: I have some doubts for (b), but a and c should be okay.
hope it helps.
 
  • #5
I pulled the formula from the formula sheet off the back of last years exam. I've done the exam now, so I'm not stressing about it, and nor was there a question about this too much on the exam I did, I was just wondering because finding a solid answer is hard.

And yes, I know how to use 2d*sin(theta)=n*lambda

b was correct, from last years answer sheet, but the answer for a and c was 'explanation required'

In the exam I just completed, there was a similar question and it hinted at K-alpha and K-beta energy differences. Is this correct? No need to be so cryptic, lol, just curious, not actually homework anymore.
 

FAQ: X-ray Diffraction, Intensity vs. Wavelength Graph

1. What is X-ray diffraction?

X-ray diffraction is a technique used to study the atomic and molecular structure of materials. It involves shining a beam of X-rays onto a sample and measuring the intensity and angle of the scattered X-rays to determine the arrangement of atoms in the sample.

2. What is an intensity vs. wavelength graph in X-ray diffraction?

An intensity vs. wavelength graph in X-ray diffraction is a plot of the intensity of the diffracted X-rays at different wavelengths. It is used to analyze the diffraction pattern produced by a sample and can provide information about the crystal structure and composition of the sample.

3. How is the intensity vs. wavelength graph obtained in X-ray diffraction?

The intensity vs. wavelength graph is obtained by measuring the intensity of the diffracted X-rays at different angles using a detector. The X-ray wavelength can also be varied using a monochromator. The data is then plotted on a graph to show the diffraction peaks and their corresponding intensities.

4. What can be learned from an intensity vs. wavelength graph in X-ray diffraction?

An intensity vs. wavelength graph can provide information about the crystal structure, lattice spacing, and composition of the sample. It can also reveal the presence of impurities or defects in the crystal lattice, as well as the orientation of the crystal relative to the incident X-ray beam.

5. How is an intensity vs. wavelength graph used in materials science and research?

An intensity vs. wavelength graph is an important tool in materials science and research. It is used to identify and characterize new materials, study the effects of processing on materials, and determine the quality and purity of materials. It is also used to compare diffraction patterns of known and unknown samples to aid in material identification and analysis.

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