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Chapter 10: Problem 50

The second-order diffraction \((n=2)\) for a gold crystal is at an angle of \(22.20^{\circ}\) for \(\mathrm{X}\) rays of 154 \(\mathrm{pm} .\) What is the spacing between these crystal planes?

Short Answer

Expert verified
The spacing between the crystal planes in the gold crystal can be found using Bragg's Law, \( n\lambda = 2d\sin \theta \). Plugging in the given values, we get \( d = \frac{2 \cdot 154\, \mathrm{pm}}{2\sin(22.20^{\circ})} \approx 200.78\, \mathrm{pm} \). Thus, the spacing between the crystal planes is approximately \(200.78\, \mathrm{pm}\).

Step by step solution

01

Understand Bragg's Law

Bragg's Law is given by \(n\lambda = 2d\sin \theta\), where: - \(n\) is the order of diffraction (in this case, \(n=2\)) - \(\lambda\) is the wavelength of the incident rays (in this case, \(154 \, \mathrm{pm}\)) - \(d\) is the spacing between the crystal planes (the quantity we want to find) - \(\theta\) is the angle of diffraction (in this case, \(22.20^{\circ}\))
02

Plug in the given values

We will plug in the given values of \(n\), \(\lambda\), and \(\theta\) into Bragg's Law equation: \[2\cdot 154 \, \mathrm{pm} = 2d\sin (22.20^{\circ})\]
03

Solve for the spacing\(d\)

Now we need to solve for the spacing between the crystal planes, \(d\). To do this, we can divide both sides of the equation by \(2\sin(22.20^{\circ})\). \[d = \frac{2\cdot 154\, \mathrm{pm}}{2\sin(22.20^{\circ})}\]
04

Calculate the final answer

Finally, we can plug in the values and calculate the spacing between the crystal planes, \(d\): \[d = \frac{2 \cdot 154\, \mathrm{pm}}{2\sin(22.20^{\circ})} \approx 200.78\, \mathrm{pm}\] The spacing between the crystal planes in the gold crystal is approximately \(200.78\, \mathrm{pm}\).

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