Problem 34 - University Physics with Modern Physics

Question

If the planes of a crystal are 3.50 \(\AA\) (1 \(\AA\) = 10\(^{-10}\) m = 1 \(\AA\)ngstrom unit) apart, (a) what wavelength of electromagnetic waves is needed so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of 22.0\(^\circ\), and in what part of the electromagnetic spectrum do these waves lie? (See Fig. 32.4.) (b) At what other angles will strong interference maxima occur?

Step-by-Step Solution

Answer

(a) Wavelength = 2.62 Å; it is in the X-ray spectrum. (b) Other angles: 48.5° for n=2.

Step 1: Identify Known Values

We have the spacing between the crystal planes, given as 3.50 Å. Convert this to meters: 3.50 Å = 3.50 × 10⁻¹⁰ m. The angle of incidence is given as 22.0°, and we need to determine the wavelength of electromagnetic waves required for the first maximum, which corresponds to n = 1 in the Bragg's Law formula.

Step 2: Apply Bragg's Law

Bragg's Law is given by the formula \(n\lambda = 2d\sin(\theta)\), where \(n\) is the order (n=1 for first maximum), \(\lambda\) is the wavelength, \(d\) is the distance between the planes, and \(\theta\) is the angle of incidence. Substituting the known values: \(1 \cdot \lambda = 2 \cdot 3.50 \times 10^{-10} \cdot \sin(22.0°)\).

Step 3: Calculate Wavelength

Solve the equation for \(\lambda\): \(\lambda = 2 \cdot 3.50 \times 10^{-10} \cdot \sin(22.0°)\). Calculate \(\sin(22.0°) = 0.3746\). Therefore, \(\lambda = 2 \cdot 3.50 \times 10^{-10} \cdot 0.3746 = 2.62 \times 10^{-10} \) m or 2.62 Å.

Step 4: Determine Electromagnetic Spectrum Part

The calculated wavelength \(\lambda = 2.62 \) Å falls in the X-ray portion of the electromagnetic spectrum, which typically ranges from 0.01 to 10 Å.

Step 5: Find Other Angles for Maximum

Strong interference maxima also occur at other angles corresponding to higher orders \(n\). Use the modified Bragg equation: \(n\lambda = 2d \sin(\theta)\), where \(n=2, 3, \) etc. For \(n=2\), solve \(2 \cdot 2.62 \times 10^{-10} = 2 \cdot 3.50 \times 10^{-10} \cdot \sin(\theta)\), giving \(\theta = \sin^{-1}(0.748) = 48.5°\). Continue similarly for other integer \(n\).

Key Concepts

Crystal Plane Spacing

In the realm of crystallography, understanding the spacing between crystal planes is crucial. These crystal planes are imagined layers of atoms or molecules inside a crystal. The spacing between these planes, often denoted by \(d\), helps in determining various properties, such as diffraction patterns. In this exercise, the crystal plane spacing is given as 3.50 Å, which equals 3.50 × 10\(^{-10}\) meters. This measurement impacts how electromagnetic waves interact within the crystal structures. Bragg's Law utilises this spacing to calculate the wavelengths that can induce constructive interference, where waves reflect off these planes. This often explains patterns seen in X-ray diffraction experiments. Smaller plane spacing typically influences higher energy regions within the electromagnetic spectrum like X-ray or even gamma rays.

Interference Maxima

The concept of interference maxima is pivotal in wave physics. It refers to the points where waves superpose to create a resultant wave of greater amplitude. In the context of Bragg's reflections, interference maxima occur when electromagnetic waves constructively interfere after reflecting off crystal planes. The angle at which these interference maxima appear depends on both the wavelength of the incident waves and the plane spacing. Bragg's Law, given by the equation \(n\lambda = 2d\sin(\theta)\), allows us to find these angles quantitatively. In this exercise, for the first maximum (n=1), we found it at an angle of 22.0° using the given parameters. This phenomenon is critical in techniques such as X-ray crystallography, where understanding these patterns helps decipher a crystal's atomic structure.

Electromagnetic Spectrum

The electromagnetic spectrum includes all types of electromagnetic waves, varying in wavelength and frequency. It encompasses regions like radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Each type of wave serves different purposes and has unique properties. In this problem, the calculated wavelength was found to be 2.62 Å, placing it within the X-ray region of the electromagnetic spectrum. This part of the spectrum is known for its penetrating nature, making it ideal for revealing internal structures through imaging techniques. Understanding where a wavelength falls in the spectrum helps in determining its applications and potential interactions with different materials.

Wavelength Calculation

Calculating the wavelength of waves required for certain phenomena is a common task in physics. With Bragg's Law, we seek to find the specific wavelengths that yield constructive interference at particular angles. This involves rearranging the equation \(n\lambda = 2d\sin(\theta)\) to solve for \(\lambda\).In this exercise, we determined \(\lambda = 2.62 \times 10^{-10}\) m or 2.62 Å, using the angle \(\theta=22.0^\circ\). This was achieved by substituting the known values into the formula and performing simple trigonometric computations. Such calculations are integral in fields like materials science, where precise measurements are necessary to understand the properties of various engineering materials and biological specimens.