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Chapter 10: Problem 48

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 of the gold crystal is approximately 198.08 pm. To find this, we used Bragg's Law \[ n\lambda = 2d\sin\theta \] with the given values of the second-order diffraction angle θ = 22.2°, and the wavelength λ = 154 pm.

Step by step solution

01

Understand Bragg's Law

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

Insert the given values into Bragg's law

We're given that the order of diffraction, n, is 2, the wavelength λ is 154 pm, and the diffraction angle θ is 22.2°. Substitute these values into Bragg's Law: \[2(154\,\mathrm{pm}) = 2d\sin(22.20^{\circ})\]
03

Solve for d

Now, we just need to isolate the variable, d, by dividing both sides of the equation by \(2\sin(22.2^{\circ})\): \[d = \frac{2(154\,\mathrm{pm})}{2\sin(22.20^{\circ})}\]
04

Calculate the value for d

Using a calculator, plug in the values and find d: \[d = \frac{2(154\,\mathrm{pm})}{2\sin(22.20^{\circ})} = 198.08\,\mathrm{pm}\]
05

Write the Final Answer

The spacing between the crystal planes of the gold crystal is approximately 198.08 pm.

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