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X rays of wavelength 0.0850 \(\mathrm{nm}\) are scattered from the atoms of a crystal. The second-order maximum in the Bragg reflection occurs when the angle \(\theta\) in Fig. 36.23 is \(21.5^{\circ} .\) What is the spacing between adjacent atomic planes in the crystal?

Short Answer

Expert verified
The spacing between adjacent atomic planes is approximately 0.232 nm.

Step by step solution

01

Write the Bragg's Law Formula

The Bragg's Law formula is used to find the spacing between crystal planes and is given by:\[ n\lambda = 2d\sin\theta \]where \( n \) is the order of the maximum, \( \lambda \) is the wavelength of the X-rays, \( d \) is the spacing between planes, and \( \theta \) is the angle of incidence/reflection.
02

Identify Known Values

We know the wavelength of the X-rays is \( \lambda = 0.0850 \, \mathrm{nm} \), the angle \( \theta = 21.5^{\circ} \), and it's the second-order maximum, so \( n = 2 \).
03

Convert Wavelength to Meters

Convert the wavelength from nanometers to meters, because standard units for calculations in physics involve meters.\[ \lambda = 0.0850 \, \mathrm{nm} = 0.0850 \times 10^{-9} \, \mathrm{m} \]
04

Solve for the Spacing \(d\)

Rearrange Bragg's Law to solve for \( d \):\[ d = \frac{n\lambda}{2\sin\theta} \]Substitute the known values:\[ d = \frac{2 \times 0.0850 \times 10^{-9} \, \text{m}}{2\sin(21.5^{\circ})} \]
05

Calculate \( \\sin(21.5^{\circ}) \)

Calculate the sine of the angle 21.5 degrees:\[ \sin(21.5^{\circ}) \approx 0.366 \]
06

Perform Final Calculation

Substitute \( \sin(21.5^{\circ}) \) into the equation:\[ d = \frac{2 \times 0.0850 \times 10^{-9} \, \text{m}}{2 \times 0.366} \]Calculate the result:\[ d \approx \frac{0.170 \times 10^{-9} \, \text{m}}{0.732} \approx 2.32 \times 10^{-10} \, \text{m} \]
07

Convert Result to Nanometers

To express the spacing \( d \) in nanometers, convert from meters to nanometers by multiplying by \( 10^{9} \):\[ d \approx 2.32 \times 10^{-10} \, \text{m} \times 10^9 = 0.232 \, \text{nm} \]

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Key Concepts

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

X-ray Scattering
X-ray scattering is a fascinating phenomenon where X-rays are deflected as they pass through a material. This occurs because X-rays, which are a type of electromagnetic radiation with very short wavelengths, interact with the atoms within a crystal. The scattering takes place due to interactions with the electrons in the atoms, resulting in an interference pattern. This pattern can be analyzed to provide valuable insights into the internal structure of the material.

One of the most significant applications of X-ray scattering is in crystallography, where it helps to determine the arrangement of atoms in a crystal. When X-rays hit a crystal, they are scattered in different directions. By measuring the angles and intensities of these scattered X-rays, scientists can infer the three-dimensional position of atoms in the crystal.

An important aspect of X-ray scattering in crystals is Bragg's Law, which links the scattering angles to the crystal lattice spacing. This relationship is crucial for identifying peaks in the scattering intensity, known as Bragg peaks, which occur when scattered waves reinforce each other. These peaks are used to deduce information about the crystal's geometry and periodicity.
Crystal Lattice Spacing
Crystal lattice spacing refers to the distance between planes of atoms in a crystal, which is a fundamental characteristic influencing the crystal's properties. In crystal structures, atoms are arranged in a repeating pattern, forming a three-dimensional lattice. The spacing between these lattice planes can be measured using techniques such as X-ray diffraction.

According to Bragg's Law, the condition for constructive interference, which results in intense diffraction peaks, is given by:
  • \[ n\lambda = 2d\sin\theta \]
Here, \( n \) indicates the order of reflection; \( \lambda \) is the X-ray wavelength; \( d \) is the crystal lattice spacing, and \( \theta \) is the angle of incidence.

This equation highlights how the lattice spacing \( d \) can be calculated if other parameters like wavelength and the reflection angle are known. When X-rays encounter planes of atoms spaced by distance \( d \), they reflect at specific angles depending on their wavelength.

Using Bragg's Law, scientists can decipher the crystal structure, which is vital in fields like material science and chemistry for understanding material properties and behaviors.
Wavelength Conversion
Wavelength conversion is an essential step in many physics calculations, especially those involving electromagnetic waves like X-rays. In scientific computations, the standard unit for length is meters, so converting wavelength measurements from other units such as nanometers is crucial.

To convert a wavelength from nanometers to meters, one must multiply by \( 10^{-9} \). For example, a wavelength of 0.0850 nm is equivalent to \( 0.0850 \times 10^{-9} \) meters.

This conversion allows for coherent calculations within Bragg’s Law and other related equations, ensuring that all terms in an equation are in the correct units. Consistency in units is vital for ensuring accurate and meaningful results in physics problems.

Moreover, understanding this fundamental conversion helps in interpreting the results of calculations within a broader scientific context, linking the nanoscale phenomena observed in crystallography to tangible distances in metric units. This skill is not only important in academic exercises but also in practical applications where precise measurements are critical.

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Most popular questions from this chapter

An interference pattern is produced by light of wavelength 580 \(\mathrm{nm}\) from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 \(\mathrm{mm}\) . (a) If the slits are very narrow, what would be the angular positions of the first-order and second- order, two-slit, interference maxima? (b) Let the slits have width 0.320 \(\mathrm{mm}\) , In terms of the intensity \(I_{0}\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

What is the longest wavelength that can be observed in the third order for a transmission grating having 6500 slits \(/ \mathrm{cm} ?\) Assume normal incidence.

Phased-Array Radar. In one common type of radar installation, a rotating antenna sweeps a radio beam around the sky. But in a phased-array radar system, the antennas remain stationary and the beam is swept electronically. To see how this is done, consider an array of \(N\) antennas that are arranged along the horizontal \(x\) -axis at \(x=0, \pm d, \pm 2 d, \ldots, \pm(N-1) d / 2 .\) (The number \(N\) is odd.) Each antenna emits radiation uniformly in all directions in the horizontal \(x y\) -plane. The antennas all emit radiation coherently, with the same amplitude \(E_{0}\) and the same wavelength \(\lambda\) . The relative phase \(\delta\) of the emission from adjacent antennas can be varied, however. If the antenna at \(x=0\) emits a signal that is given by \(E_{0} \cos \omega t,\) as measured at a point next to the antenna, the antenna at \(x=d\) emits a signal given by \(E_{0} \cos (\omega t+\delta),\) as measured at a point next to that antenna. The corresponding quantity for the antenna at \(x=-d\) is \(E_{0} \cos (\omega t-\delta) ;\) for the antennas at \(x=\pm 2 d,\) it is \(E_{0} \cos (\omega t \pm 2 \delta) ;\) and so on. \((a)\) If \(\delta=0\) , the interference pattern at a distance from the antennas is large compared to \(d\) and has a principal maximum at \(\theta=0\) (that is, in the angular range \(-90^{\circ}<\theta<90^{\circ} .\) Hence this principal maximum describes a beam emitted in the direction \(\theta=0\) . As described in Section \(36.4,\) if \(N\) is large, the beam will have a large intensity and be quite narrow. (b) If \(\delta \neq 0\) , show that the principal intensity maximum described in part (a) is located at $$ \boldsymbol{\theta}=\arcsin \left(\frac{\delta \boldsymbol{\lambda}}{2 \pi d}\right) $$ where \(\delta\) is measured in radians. Thus, by varying \(\delta\) from positive to negative values and back again, which can easily be done electronically, the bearn can be made to sweep back and forth around \(\boldsymbol{\theta}=\mathbf{0} .\) (c) A weather radar unit to be installed on an airplane emits radio waves at 8800 \(\mathrm{MHz}\) . The unit uses 15 antennas in an array 28.0 \(\mathrm{cm}\) long (from the antenna at one end of the array to the antenna at the other end). What must the maximum and minimum values of \(\delta\) be (that is, the most positive and most negative values) if the radar beam is to sweep \(45^{\circ}\) to the left or right of the air- plane's direction of flight? Give your answer in radians.

(a) What is the wavelength of light that is deviated in the first order through an angle of \(13.5^{\circ}\) by a transmission grating having 5000 slits \(/ \mathrm{cm} ?\) (b) What is the second order deviation of this wavelength? Assume normal incidence.

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.

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