Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 48

The second-order diffraction \((n=2)\) for a gold crystal is at an angle of \(22.20^{\circ}\) for \(\mathrm{X}\) rays of \(154 \mathrm{pm}\). What is the spacing between these crystal planes?

Short Answer

Expert verified
The spacing between the crystal planes in the gold crystal can be calculated using Bragg's Law formula: \(n\lambda = 2d \sin{\theta}\). Given the second-order diffraction (n=2), X-ray wavelength \(\lambda = 154\,\text{pm}\), and angle \(\theta = 22.20^{\circ}\), we can find the interplanar spacing (d) by rearranging the formula to \(d = \frac{n\lambda}{2\sin{\theta}}\). Substituting the given values, we get \(d \approx \frac{308\,\text{pm}}{\sin{22.20^{\circ}}} \approx 797.76\,\text{pm}\). Therefore, the spacing between the crystal planes is approximately \(797.76\,\text{pm}\).

Step by step solution

01

Recall the Bragg's Law Formula

Bragg's Law states that for constructive interference to occur, the path difference between two rays reflecting off adjacent planes should be an integer multiple of the wavelength. The formula is given by: \(n\lambda = 2d \sin{\theta}\) where n = order of diffraction (integer), λ = wavelength of the incident ray, d = interplanar spacing (the distance between adjacent crystal planes), θ = angle of incidence/reflection.
02

Substitute the given values

We have been given the following data: n = 2 (second-order diffraction), λ = 154 pm, θ = 22.20 degrees. Now, put these values in Bragg's Law formula: \(2(154\,\text{pm}) = 2d \sin{22.20^{\circ}}\)
03

Solve for interplanar spacing (d)

Now, we need to find d by rearranging the formula and then solving it: \(d = \frac{2(154\,\text{pm})}{2\sin{22.20^{\circ}}}\) Which simplifies to: \(d = \frac{308\,\text{pm}}{\sin{22.20^{\circ}}}\) Now, find the value of d: \(d = \frac{308\,\text{pm}}{\sin{22.20^{\circ}}} \approx 797.76\,\text{pm}\)
04

Round the result and state the answer

After rounding the result to two decimal places, the spacing between these crystal planes is: \(d \approx 797.76\,\text{pm}\) Thus, the spacing between the crystal planes in the gold crystal is approximately 797.76 pm.

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.

Start your free trial

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

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

A certain metal fluoride crystallizes in such a way that the fluoride ions occupy simple cubic lattice sites, while the metal ions occupy the body centers of half the cubes. What is the formula of the metal fluoride?

A 0.132 -mole sample of an unknown semiconducting material with the formula XY has a mass of \(19.0 \mathrm{g}\). The element X has an electron configuration of \([\mathrm{Kr}] 5 s^{2} 4 d^{10} .\) What is this semiconducting material? A small amount of the Y atoms in the semiconductor is replaced with an equivalent amount of atoms with an electron configuration of \([\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{5} .\) Does this correspond to n-type or p-type doping?

Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of \(197 \mathrm{pm},\) calculate the density of solid calcium.

Carbon tetrachloride, \(\mathrm{CCl}_{4},\) has a vapor pressure of 213 torr at \(40 .^{\circ} \mathrm{C}\) and 836 torr at \(80 .^{\circ} \mathrm{C} .\) What is the normal boiling point of \(\mathrm{CCl}_{4} ?\)

Is it possible for the dispersion forces in a particular substance to be stronger than the hydrogen bonding forces in another substance? Explain your answer.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation
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