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Chapter 10: Problem 46

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

Short Answer

Expert verified
The spacing between the crystal planes is approximately 402.70 pm.

Step by step solution

01

Write down Bragg's Law

Bragg's Law is given by the formula: \[n\lambda = 2d\sin{\theta}\] Where: - \(n\) is the order of diffraction, - \(\lambda\) is the wavelength of the X-rays, - \(d\) is the spacing between the crystal planes, and - \(\theta\) is the angle of diffraction.
02

Plug in the given values into Bragg's Law

We have been given the following values: - \(n = 2\) (second-order diffraction), - \(\lambda = 154 \, \text{pm}\) (wavelength of the X-rays), and - \(\theta = 22.20^\circ\) (angle of diffraction). Now substitute these values in Bragg's Law: \(2(154 \, \text{pm}) = 2d \sin{22.20^\circ}\)
03

Solve the equation for the spacing between the crystal planes (d)

To find the spacing between the crystal planes, \(d\), we now just need to solve the equation from Step 2: \(308 \, \text{pm} = 2d \sin{22.20^\circ}\) Next, divide both sides of the equation by \(2\sin{22.20^\circ}\): \(d = \frac{308 \, \text{pm}}{2 \sin{22.20^\circ}}\) Finally, compute the value of \(d\): \(d \approx 402.70 \, \text{pm}\) The spacing between the crystal planes is approximately 402.70 pm.

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