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Chapter 28: Problem 10

Neutron diffraction by a crystal can be used to make a velocity selector for neutrons. Suppose the spacing between the relevant planes in the crystal is \(d=0.20 \mathrm{nm}\) A beam of neutrons is incident at an angle \(\theta=10.0^{\circ}\) with respect to the planes. The incident neutrons have speeds ranging from 0 to \(2.0 \times 10^{4} \quad \mathrm{m} / \mathrm{s} .\) (a) What wavelength(s) are strongly reflected from these planes? [Hint: Bragg's law, Eq. \((25-15),\) applies to neutron diffraction as well as to x-ray diffraction.] (b) For each of the wavelength(s), at what angle with respect to the incident beam do those neutrons emerge from the crystal?

Short Answer

Expert verified
2. What are the given values for d and 胃 in this exercise? 3. Which steps should be followed to solve this exercise? 4. What is the angle of emergence for the calculated wavelength(s)?

Step by step solution

01

Calculate the Wavelength(s)

First, let's calculate the wavelength 位 using Bragg's law. Since the integer n can take any value, we need to find the minimum value of n that will give a meaningful result. Bragg's law: n位 = 2d sin胃 Rearrange to find 位: 位 = (2d sin胃)/n
02

Calculate the Angles

Now, let's calculate the angle at which the neutrons emerge from the crystal for each of the wavelength(s) found in step 1. We know that the angle of emergence will be equal to the angle of incidence, 胃. So, the angle of emergence with respect to the incident beam will be 胃. In this case, 胃 = 10.0掳.

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