The first order diffraction of \(X\) -rays from a certain set of crystal planes oceurs at an angle of \(11.8^{\circ}\) from the planes. If the planes are \(0.281 \mathrm{~nm}\) apart, what is the wavelength of \(X\) -rays?

Short Answer

Expert verified
The wavelength of the X-rays is approximately \( 0.0579 \text{ nm} \).

Step by step solution

01

Understanding Bragg's Law

Recognize that the problem involves X-ray diffraction and Bragg's Law which relates the angle of incidence, the distance between crystal planes, and the wavelength of light. Bragg's Law is given by the equation \( n\lambda = 2d\sin(\theta) \), where \( n \) is the order of diffraction, \( \lambda \) is the wavelength, \( d \) is the distance between crystal planes, and \( \theta \) is the angle of diffraction.
02

Input Given Values

Since the first-order diffraction is given, set \( n = 1 \). The angle of diffraction \( \theta \) is given as \( 11.8^\circ \) and the crystal plane separation \( d \) is \( 0.281 \) nm. Convert \( d \) to meters by multiplying by \( 10^{-9} \) to get \( 0.281 \times 10^{-9} \) meters.
03

Solve for the Wavelength

Using Bragg's Law, solve for the wavelength \( \lambda \) by rearranging the equation to \( \lambda = \frac{2d\sin(\theta)}{n} \). Substitute the known values \( d = 0.281 \times 10^{-9} \) meters, \( \theta = 11.8^\circ \) (convert to radians for calculation), and \( n = 1 \), then calculate \( \lambda \).
04

Conversion of Angle into Radians

Convert the angle from degrees to radians by using the conversion factor \( \pi \) radians = \( 180^\circ \). Therefore, \( \theta = 11.8^\circ = 11.8 \times \frac{\pi}{180} \) radians.
05

Calculate the Wavelength

Perform the calculation using the converted angle: \( \lambda = 2 \times 0.281 \times 10^{-9} \times \sin(11.8 \times \frac{\pi}{180}) \), which yields the wavelength in meters. Lastly, convert the wavelength to nanometers if required by multiplying the result by \( 10^9 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding X-ray Diffraction
X-ray diffraction (XRD) is a powerful tool widely used to analyze the structure of crystalline materials. At its core, XRD is based on the phenomenon that occurs when X-rays are shined on a crystal, causing them to scatter in many directions. However, constructive interference (when the waves amplify each other) only happens at specific angles, determined by the spacing between the crystal planes and the X-ray's wavelength. This is known as Bragg's Law. When X-rays hit the planes within the crystal at the correct angle, they are diffracted, and this diffraction pattern can be measured.

XRD is crucial for determining various properties of the material, such as lattice parameters, crystal structure, and even the presence of defects or impurities within the crystal. As students delve into the world of crystallography and materials science, understanding the principles of X-ray diffraction becomes a foundational aspect of their studies.
Exploring Crystal Planes
Crystal planes are, in effect, imaginary planes that pass through crystal lattice points. Crystals have a periodic structure, and the orientation of these planes is critical for understanding X-ray diffraction. Imagine crystal planes as the slices in a loaf of bread, with each slice having a specific orientation and spacing.

These planes are defined by three indices, known as Miller indices, which denote the orientation of the planes in the crystal structure. The distance between these planes, denoted as 'd' in Bragg's Law, affects the path and hence the diffraction pattern of the X-rays passing through them. The closer the planes are, the higher the angle required for X-rays to meet the conditions of Bragg's Law. Scientists and engineers utilize the concept of crystal planes to interpret the XRD data and gain insights into the material's structure.
Calculating the Wavelength with Bragg's Law
Wavelength calculation is an integral part of XRD analysis and is systematically executed using Bragg's Law. Let's break down the steps to derive the wavelength from the given data. When the first-order diffraction pattern is observed at an angle, you have the variables needed to solve for the X-ray's wavelength. The law's formula,
\[\begin{equation} n\textbackslashlambda = 2d\textbackslashsin(\textbackslashtheta) \end{equation}\]

provides a clear equation to follow, where the known diffraction order (n), crystal plane spacing (d), and diffraction angle (θ) can be plugged in to solve for the unknown wavelength (λ). The exercise involves converting units and angles into compatible formats for calculation. Angles are converted to radians for this purpose, as it's a requirement of the sine function when using the formula. After the variables are substituted, the rearranged Bragg's Law equation allows for the computation of the wavelength. This calculated wavelength is significant because it reveals details about the X-ray source and can also be used to further interpret the crystal structure.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

State which one is homogeneous or heterogeneous? (a) \(\quad \overline{\text { C }_{\text {Diamond }}} \rightleftharpoons \bar{C}_{\text {graphite }}\) (b) \(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})}=\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\) (c) \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}\) (d) \(\quad \mathrm{MgCO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{MgO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{~g})}\) (e) \(\mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}=\mathrm{PCl}_{5(\mathrm{~g})}\)

\(100 \mathrm{~g}\) of \(\mathrm{NaCl}\) is stirred in \(100 \mathrm{~mL}\) of water at \(20^{\circ} \mathrm{C}\) till the equilibrium is attained : (a) How much \(\mathrm{NaCl}\) goes into the solution and how much of it is left undissolved at equilibrium? The solubility of \(\mathrm{NaCl}\) at \(20^{\circ} \mathrm{C}\) is \(6.15\) mol/litre. (b) What will be the amount of \(\mathrm{NaCl}\) left undissolved if the solution is diluted to \(200 \mathrm{~mL} ?\)

For the gaseous reaction; \(2 \mathrm{NO}_{2} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}\), calculate \(\Delta \mathrm{G}^{\circ}\) and \(K_{\mathrm{p}}\) for the reaction at \(25^{\circ} \mathrm{C}\). Given \(\mathrm{G}_{j}^{\circ} \mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{G}_{f}^{\circ} \mathrm{NO}_{2}\) are \(97.82\) and \(51.30\) \(\mathrm{kJ}\) respectively. Also calculate \(\Delta \mathrm{G}^{\circ}\) and \(K_{\mathrm{p}}\) for reverse reaction.

If a mixture of 3 moles of \(\mathrm{H}_{2}\) and one mole of \(\mathrm{N}_{2}\) is completely converted into \(\mathrm{NH}_{3}\). What would be the ratio of the initial and final volume at same temperature and pressure?

Which of the following reactions will get affected by increase of pressure? Also mention, whether change will cause the reaction to go into the right or left direction? (i) \(\mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{~S}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CS}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{~S}_{(\mathrm{g})}\) (ii) \(\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{C}_{(\mathrm{s})} \rightleftharpoons 2 \mathrm{CO}_{(\mathrm{g})}\) (iii) \(4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\) (iv) \(\mathrm{C}_{2} \mathrm{H}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6(\mathrm{~g})}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free