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Chapter 10: Problem 47

\(X\) rays from a copper X-ray tube \((\lambda=154 \mathrm{pm})\) were diffracted at an angle of \(14.22\) degrees by a crystal of silicon. Assuming first-order diffraction \((n=1\) in the Bragg equation), what is the interplanar spacing in silicon?

Short Answer

Expert verified
The interplanar spacing in the silicon crystal is approximately \(191.5 \, \text{pm}\).

Step by step solution

01

Write down Bragg's Law

Bragg's Law relates the X-ray wavelength, diffraction angle, order of diffraction, and interplanar spacing. It is given by: \(2 d \sin{\theta}=n\lambda\) where, - \(d\) is the interplanar spacing - \(\theta\) is the diffraction angle - \(n\) is the order of diffraction - \(\lambda\) is the X-ray wavelength.
02

Write down the given values

We are given the following values: - X-ray wavelength, \(\lambda = 154 \, \text{pm}\) - Diffraction angle, \(\theta = 14.22^{\circ}\) - Order of diffraction, \(n = 1\)
03

Convert the angle to radians

In order to use Bragg's Law, we need to convert the given angle in degrees to radians. The conversion formula is: \(\theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180}\) Therefore, we can calculate the angle in radians as follows: \(\theta_{\text{radians}} = 14.22^{\circ} \times \frac{\pi}{180} \approx 0.248 \, \text{radians}\)
04

Use Bragg's Law to find the interplanar spacing

Now, we can use Bragg's Law to find the interplanar spacing, using the given values: \(2d \sin{\theta_{\text{radians}}} = n\lambda\) Solving for \(d\): \(d = \frac{n\lambda}{2\sin{\theta_{\text{radians}}}}\) Plug in the values: \(d = \frac{1 \times 154 \, \text{pm}}{2\sin{0.248 \, \text{radians}}}\) Calculate \(d\): \(d \approx 191.5 \, \text{pm}\)
05

Write the final answer

The interplanar spacing in the silicon crystal is approximately \(191.5 \, \text{pm}\).

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