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Chapter 10: Problem 51

A topaz crystal has an interplanar spacing \((d)\) of 1.36 \(\mathrm{A}\) $\left(1 \mathrm{A}=1 \times 10^{-10} \mathrm{m}\right) .$ Calculate the wavelength of the \(\mathrm{X}\) ray that should be used if $\theta=15.0^{\circ}(\text { assume } n=1)$ .

Short Answer

Expert verified
The wavelength of the X-ray that should be used is approximately \(7.05 \times 10^{-11}\) meters.

Step by step solution

01

Convert the given interplanar spacing into meters

We know that 1 Angstrom is equivalent to \(1 \times 10^{-10}\) meters. The interplanar spacing of the topaz crystal is given in Angstroms (1.36 Å). So, we first need to convert it into meters. Interplanar spacing (d) = 1.36 Å × \(1 \times 10^{-10}\) m/Å = 1.36 × \(10^{-10}\) meters
02

Use Bragg's Law and the given values to find the wavelength

Bragg's Law is given by: \(n\lambda = 2d\sin\theta\) Substitute the given values and the value of d (in meters) into the formula: (1) × λ = 2 × (1.36 × \(10^{-10}\)) × sin(15.0°)
03

Solve for the wavelength λ

Now we can solve for λ: λ = 2 × (1.36 × \(10^{-10}\)) × sin(15.0°) Using a calculator: λ = 2 × (1.36 × \(10^{-10}\)) × 0.2588 λ ≈ 7.05 × \(10^{-11}\) meters So, the wavelength of the X-ray that should be used is approximately 7.05 × \(10^{-11}\) meters.

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Most popular questions from this chapter

Identify the most important types of interparticle forces present in the solids of each of the following substances a. Ar b. \(\mathrm{HCl}\)l c. \(\mathrm{HF}\) d. \(\mathrm{CaCl}_{2}\) e. \(C \mathrm{H}_{4}\) f. \(C O\) g. \(N a N O_{3}\)

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