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Chapter 12: Problem 120

In their study of X-ray diffraction, William and Lawrence Bragg determined that the relationship among the wavelength of the radiation \((\lambda),\) the angle at which the raveation is diffracted \((\theta),\) and the distance between planes of atoms in the crystal that cause the diffraction \((d)\) is given by \(n \lambda=2 d \sin \theta . X\) rays from a copper \(X\) -ray tube that have a wavelength of 1.54\(\hat{\mathrm{A}}\) are diffracted at an angle of 14.22 degrees by crystalline silicon. Using the Bragg equation, calculate the distance between the planes of atoms responsible for diffraction in this crystal, assuming \(n=1\) (first-order diffraction).

Short Answer

Expert verified
The distance between the planes of atoms responsible for diffraction in the silicon crystal is approximately \(3.15 \times 10^{-10} m\) or \(3.15 \mathring{A}\).

Step by step solution

01

Write down the given values

We are given the following values: - Wavelength of the radiation \(\lambda = 1.54 \mathring{A}\) (1 angstrom \(\mathring{A}\) = \(1 \times 10^{-10} m\)) - Angle of diffraction \(\theta = 14.22\) degrees - Order of diffraction \(n = 1\)
02

Substitute the given values into the Bragg equation

Now, we will substitute these values into the Bragg equation: \(n\lambda = 2d \sin \theta\) \(1 \times (1.54 \times 10^{-10}) = 2d \sin(14.22)\)
03

Solve for the distance between the planes of atoms, \(d\)

We want to solve for \(d\). To do this, we can divide both sides of the equation by \(2\sin(14.22)\): \(d = \frac{1.54 \times 10^{-10}}{2 \sin(14.22)}\) Now, we can plug in the values and find \(d\): \(d \approx \frac{1.54 \times 10^{-10}}{2 \times 0.2444} \approx 3.15 \times 10^{-10} m \) So, the distance between the planes of atoms responsible for diffraction in the silicon crystal is approximately \(3.15 \times 10^{-10} m\) or \(3.15 \mathring{A}\).

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