Explanation 1) Concept: Using Bragg’s law and the given condition, we can find the angle θ . Then, by studying F i g . 36 - 28 , we can find the angle between the incident and scattered beams. 2) Formulas: i) From E q . 36 - 34 , Bragg’s law gives 2 d s i n θ = m λ ii) From F i g . 36 - 28 , the angle θ ' between the incident and scattered beams is θ ' = 180 ° - 2 θ 3) Given: i) The range of x ray wavelengths is from 0.120 n m to 0.0700 n m . ii) The plane separation is d = 0.25 n m iii) The scattered beams are produced for λ = 0.10 n m and λ ' = 0.075 n m . 4) Calculations: From E q . 36 - 34 , Bragg’s law gives 2 d s i n θ = m λ But, for λ = 0.10 n m , we have scattering for order m , and for λ ' = 0.075 n m , we have scattering for order m ' . Therefore, we must require m λ = m ' λ ' For given values of λ a n d λ ' , this suggests that m = 3 and m ' = 4

10th Edition

ISBN: 9781118230718

Author: David Halliday

Publisher: Wiley, John & Sons, Incorporated

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Chapter 36, Problem 104P

To determine

To find:

Angle between the incident and scattered beams

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Chapter 36, Problem 103P

Chapter 36, Problem 105P

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Chapter 36 Solutions

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Angle between the incident and scattered beams

Solution Summary: The author explains how to find the angle between the incident and scattered beams using Bragg's law and the given condition.

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To find: Angle between the incident and scattered beams

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Chapter 36 Solutions

To find:

Angle between the incident and scattered beams