Explanation 1) Concept: By using Bragg’s diffraction condition and from the given graph, we can find the shorter and the longer wavelength in the beam. 2) Formula: 2 d sin ⁡ θ = m λ 3) Given: i) Interplanar spacing, d = 0.94 n m ii) Horizontal scale, θ s = 2.00 ° iii) Order, m = 1 4) Calculations: a) By considering the graph from Fig. 36-46 and using the given horizontal scale, we can estimate that the first peak is at an angle 0.75 ° . So by Bragg’s diffraction condition for the first order, we can find the shorter wavelength of the beam. 2 d sin ⁡ θ 1 = λ 1 λ 1 = 2 × 0.94 × sin ⁡ 0.75 0 λ 1 = 2 × 0

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Chapter 36, Problem 67P

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To find:

a) The shorter wavelength in the beam

b) The longer wavelength in the beam

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Chapter 36, Problem 66P

Chapter 36, Problem 68P

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Chapter 36 Solutions

Fundamentals of Physics Extended

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a) The shorter wavelength in the beam b) The longer wavelength in the beam

Solution Summary: The author explains Bragg's diftion condition and the given graph to find the shorter and longer wavelengths in the beam.

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To find: a) The shorter wavelength in the beam b) The longer wavelength in the beam

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Chapter 36 Solutions

To find:

a) The shorter wavelength in the beam

b) The longer wavelength in the beam