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Chapter 36: Problem 40

If the planes of a crystal are 3.50 \(\mathrm{A}\left(\mathrm{i} \mathrm{A}=10^{-10} \mathrm{m}=\right.\) 1 Angstrom unit) apart, (a) what wavelength of electromagnetic waves is necded so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of \(15.0^{\circ}\) , 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?

Short Answer

Expert verified
(a) Wavelength is 1.81 脜 in X-rays. (b) Other strong maxima at approximately 31.1 degrees.

Step by step solution

01

Understand Bragg's Law

Bragg's Law is given by:\[ n\lambda = 2d \sin\theta \]where \( n \) is the order of the maximum, \( \lambda \) is the wavelength, \( d \) is the distance between planes, and \( \theta \) is the angle of incidence. To find the first interference maximum, we set \( n = 1 \). This will give us the wavelength for the first-order maximum.
02

Solve for Wavelength (\(\lambda\))

Using Bragg's Law for the first-order interference maximum (\(n = 1\)) and the given values:\[ \lambda = \frac{2d \sin \theta}{n} \]\[ \lambda = \frac{2(3.50 \, \text{脜}) \sin(15.0^\circ)}{1} \]First convert 3.50 脜 to meters.\[ 3.50 \, \text{脜} = 3.50 \times 10^{-10} \, \text{m} \]Now, calculate \(\sin(15.0^\circ)\), which is approximately 0.2588.Substitute these into the equation:\[ \lambda = 2 \times 3.50 \times 10^{-10} \, \text{m} \times 0.2588 \]\[ \lambda \approx 1.81 \times 10^{-10} \, \text{m} \]
03

Determine Electromagnetic Spectrum Region

The wavelength \( \lambda \approx 1.81 \times 10^{-10} \, \text{m} \) corresponds to X-rays in the electromagnetic spectrum. X-rays typically have wavelengths ranging from about 0.01 nm (or 0.1 脜) to 10 nm (or 100 脜).
04

Find Other Angles for Interference Maxima

For higher-order interference maxima, use the generalized form of Bragg's Law:\[ n\lambda = 2d \sin\theta \]For the second maximum (\(n = 2\)), solve:\[ n\lambda = 2d \sin\theta \]\[ 2(1.81 \times 10^{-10}) = 2 (3.50 \times 10^{-10} \sin \theta) \]\[ \sin \theta = \frac{2(1.81 \times 10^{-10})}{2(3.50 \times 10^{-10})} \]\[ \sin \theta \approx 0.5171 \]\[ \theta \approx 31.1^\circ \]Check if higher orders are possible by ensuring \( \sin \theta \leq 1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electromagnetic Spectrum
The electromagnetic spectrum is a vast range of wavelengths that encompass all types of electromagnetic radiation. Spanning from the longest wavelengths such as radio waves, to the shortest like gamma rays, the spectrum reveals how varied and dynamic electromagnetic energy can be. It is generally divided into several regions according to wavelength and frequency:
  • Radio Waves
  • Microwaves
  • Infrared
  • Visible Light
  • Ultraviolet
  • X-rays
  • Gamma Rays
Each section has its unique characteristics and applications. Radio waves are used in communication, while visible light is what we can see with our eyes.
X-rays and gamma rays, found at the shorter wavelength end, are known for their penetrating power and are typically used in medical imaging and treatments. Understanding where certain waves fall on the electromagnetic spectrum can greatly help in the application of these energies in various fields.
X-rays
X-rays are a type of electromagnetic radiation with very short wavelengths. They range between 0.01 to 10 nanometers (or 0.1 and 100 脜ngstroms), placing them just beyond ultraviolet light on the electromagnetic spectrum. Because of their short wavelengths, X-rays have high energy and the ability to pass through many materials, making them invaluable in medical diagnostics.
In a medical setting, they are commonly used to image bones and teeth, as their penetrative ability allows them to provide a clear picture of the denser structures in the body. X-rays are also vital in other settings such as security scanners and crystallography, where they are used to analyze the internal structure of materials at the atomic level.
Although incredibly useful, it is important to manage their use carefully due to their ionizing potential, which can lead to tissue damage with excessive exposure.
Interference Maxima
Interference maxima is a concept central to understanding wave behavior when encountering obstacles like crystal planes. In the context of X-rays hitting a crystal, it refers to the positions where waves reinforce each other to produce maximum intensity. This happens when the path differences between waves reflecting from different layers match the integer multiples of the wavelength.
Bragg's Law, given by the equation \( n\lambda = 2d \sin\theta \), is the key relationship used to predict these maxima. In it, \( n \) represents the order of the maximum, \( \lambda \) is the wavelength, \( d \) is the separation between crystal planes, and \( \theta \) is the glancing angle of incidence.
By adjusting the angle \( \theta \) or the order number \( n \), one can determine the angles at which constructive interference occurs, leading to the strongest diffraction peaks. This principle is widely employed in X-ray diffraction techniques to probe the crystalline structures of materials, unlocking insights into their molecular arrangements.

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Most popular questions from this chapter

A diffraction grating has 650 slits \(/ \mathrm{mm}\) . What is the highest order that contains the entire visible spectrum? (The wavelength range of the visible spectrum is approximately \(400-700 \mathrm{nm} . )\)

Hubble Versus Arecibo. The Hubble Space Telescope has an aperture of 2.4 \(\mathrm{m}\) and focuses visible light \((400-700 \mathrm{nm}) .\) The Arecibo radio telescope in Puerto Rico is 305 \(\mathrm{m}(1000 \mathrm{ft})\) in diameter (it is built in a mountain valley) and focuses radio waves of wavelength \(75 \mathrm{cm} .\) (a) Under optimal viewing conditions, what is the smallest crater that each of these telescopes could resolve on our moon? (b) If the Hubble Space Telescope were to be converted to surveillance use, what is the highest orbit above the surface of the earth it could have and still be able to resolve the license plate (not the letters, just the plate) of a car on the ground? Assume optimal viewing conditions, so that the resolution is diffraction limited.

Intensity Pattern of \(N\) Slits. (a) Consider an arrangement of \(N\) slits with a distance \(d\) between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda\) . Show that at a time \(t,\) the electric field at a distant point \(P\) is $$ \begin{aligned} E_{P}(t)=& E_{0} \cos (k R-\omega t)+E_{0} \cos (k R-\omega t+\phi) \\ &+E_{0} \cos (k R-\omega t+2 \phi)+\cdots \\ &+E_{0} \cos (k R-\omega t+(N-1) \phi) \end{aligned} $$ where \(E_{0}\) is the amplitude at \(P\) of the electric field due to an individual slit, \(\phi=(2 \pi d \sin \theta) / \lambda, \theta\) is the angle of the rays reaching \(P(\text { as measured from the perpendicular bisector of the slit arrangement }\)), and \(R\) is the distance from \(P\) to the most distant slit. In this problem, assume that \(R\) is much larger than \(d .\) (b) To carry out the sum in part \((a),\) it is convenient to use the complex-number relationship $$ e^{i k}=\cos z+i \sin z $$ where \(i=\sqrt{-1}\) . In this expression, \(\cos z\) is the real part of the complex number \(e^{i k},\) and \(\sin z\) is its imaginary part. Show that the electric field \(E_{P}(t)\) is equal to the real part of the complex quantity $$ \sum_{n=0}^{N-1} E_{0} e^{i(k R-\omega+n \phi)} $$ (c) Using the properties of the exponential function that \(e^{A} e^{B}=e^{(A+B)}\) and \(\left(e^{A}\right)^{n}=e^{n A},\) show that the sum in part \((b)\) can be written as $$ E_{0}\left(\frac{e^{i N \phi}-1}{e^{i \phi}-1}\right) e^{i(k R-\omega t)}=E_{0}\left(\frac{e^{i N \phi / 2}-e^{-i N \phi / 2}}{e^{i \phi / 2}-e^{-i \phi / 2}}\right) e^{i(k R-\omega t+(N-1) \phi / 2]} $$ Then, using the relationship \(e^{i k}=\cos z+i \sin z,\) show that the (real) electric field at point \(P\) is $$ E_{p}(t)=\left[E_{0} \frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right] \cos [k R-\omega t+(N-1) \phi / 2] $$ The quantity in the first square brackets in this expression is the amphitude of the electric field at \(P .\) (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle \(\theta\) is $$ I=I_{0}\left[\frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right]^{2} $$ where \(I_{0}\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case \(N=2\) . It will help to recall that \(\sin 2 A=2 \sin A \cos A\) . Explain why your result differs from Eq. \((35.10),\) the expression for the intensity in two-source interference, by a factor of \(4 .\) (Hint: Is \(I_{0}\) defined in the same way in both expressions?)

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 \(\mathrm{mm}\) wide. At the point in the pattern \(3.25^{\circ}\) from the center of the central maximum, the total phase difference between wavelets from the tup and bottom of the slit is 56.0 rad. (a) What is the wavelength of the radiation? (b) What is the intensity at this point, if the intensity at the center of the central maximum is \(I_{0} ?\)

If a diffraction grating produces its third-order bright band at an angle of \(78.4^{\circ}\) for light of wavelength 681 \(\mathrm{nm}\) , find (a) the number of slits per centimeter for the grating and \((b)\) the angular location of the first-order and second-order bright bands. (c) Will there be a fourth-order bright band? Explain.
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