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Chapter 10: Problem 49

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

Short Answer

Expert verified
The interplanar spacing in the silicon crystal is approximately \(314 \times 10^{-12} m\) or \(314 \mathrm{pm}\).

Step by step solution

01

Understand Bragg's Law

Bragg's Law relates the wavelength of the X-rays (\(\lambda\)), the diffraction angle (\(\theta\)), the interplanar spacing in the crystal (d), and the diffraction order (n). It is given by the formula: \[n\lambda = 2d \sin\theta\] In this exercise, we are given the wavelength (\(\lambda = 154 pm\)), the diffraction angle (\(\theta = 14.22^\circ\)), and are told to assume first-order diffraction (n = 1). Our task is to find the interplanar spacing (d).
02

Plug in Known Values

Now, we can plug in the known values into Bragg's Law equation: \[1(154 \times 10^{-12} m) = 2d \sin(14.22^\circ)\]
03

Solve for the Interplanar Spacing (d)

Next, we want to isolate the variable d in the equation. To do this, let's first calculate the value of \(\sin(14.22^\circ)\): \[\sin(14.22^\circ) \approx 0.2458\] Now we can substitute this value back into the equation: \[154 \times 10^{-12} m = 2d (0.2458)\] Since we want to find the value of d, we can divide both sides of the equation by \(2(0.2458)\): \[d = \frac{154 \times 10^{-12} m}{2(0.2458)}\] Finally, we can calculate the value of d: \[d \approx 314 \times 10^{-12} m\]
04

Report the Answer

The interplanar spacing in the silicon crystal is approximately \(314 \times 10^{-12} m\) or \(314 \mathrm{pm}\).

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