Bragg diffraction form an “inclined” crystal plane

Click For Summary
SUMMARY

The discussion centers on the application of Bragg's law to determine the unit cell size \( a_0 \) of a crystal when subjected to X-ray diffraction. The wavelength of the X-ray beam is specified as \( 0.260 \, \text{nm} \) and the angle of incidence \( \theta \) is \( 63.8° \). The participant initially proposed using an adjusted angle \( \alpha = \theta - 45° = 18.8° \) in their calculations, leading to the formula \( a_0 = \sqrt{2} \frac{\lambda}{2 \sin \alpha} \). However, they later discovered that the solution manual utilized the original angle \( \theta = 63.8° \) in Bragg's law, prompting a need for verification of their calculations.

PREREQUISITES
  • Understanding of Bragg's law and its application in X-ray diffraction.
  • Familiarity with crystal structures and unit cell dimensions.
  • Knowledge of trigonometric functions, particularly sine.
  • Basic principles of X-ray physics and wavelength measurements.
NEXT STEPS
  • Review the derivation and applications of Bragg's law in crystallography.
  • Explore the concept of unit cell dimensions in different crystal systems.
  • Study the impact of angle adjustments in diffraction calculations.
  • Investigate common discrepancies between theoretical calculations and solution manuals in physics problems.
USEFUL FOR

Students and professionals in physics, materials science, and crystallography who are involved in X-ray diffraction analysis and crystal structure determination.

crick
Messages
39
Reaction score
4

Homework Statement


In picture, first-order reflection from the reflection planes shown occurs when an x-ray beam of wavelength ##0.260 nm## makes an angle ##\theta=63.8°## with the top face of the crystal. What is the unit cell size ##a_0##?

Immagindfde.png


Homework Equations


Bragg law
$$d=\frac{ n \lambda} {2 \sin \Theta}$$

The Attempt at a Solution


I thought I must consider an angle ##\alpha =\theta -45°=18.8°## in bragg law and then $$a_0=\sqrt{2} \frac{ \lambda} {2 \sin \alpha}$$

That's because I was sure that the angle of diffraction ##\theta## in bragg's law must always be measured with respect to the plane as in the following picture

Immagine8888.png


Nevertheless I cheked on solution manual found online here : http://omega.altervista.org/extra/key%20to%20Physics.pdf

And I found the following solution:
Immagine33333.png

So the angle used in Bragg law is still ##\theta =63.8°## ! How can it be correct?
 

Attachments

  • Immagindfde.png
    Immagindfde.png
    1.9 KB · Views: 800
  • Immagine8888.png
    Immagine8888.png
    5.6 KB · Views: 727
  • Immagine33333.png
    Immagine33333.png
    14.3 KB · Views: 668
Physics news on Phys.org
I think you got it right and the book's solution is incorrect, but I would need to double-check my result to be sure.
 
  • Like
Likes   Reactions: crick

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
8K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
7
Views
6K