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Chapter 10: Problem 18

Will a crystalline solid or an amorphous solid give a simpler X-ray diffraction pattern? Why?

Short Answer

Expert verified
A crystalline solid will produce a simpler X-ray diffraction pattern compared to an amorphous solid. This is because crystalline solids have a well-ordered and repeating arrangement of particles, leading to a regular interference pattern with discrete and sharp bright spots (Bragg peaks). In contrast, amorphous solids have a disordered arrangement of particles, which results in a diffused, broad, and continuous interference pattern.

Step by step solution

01

Understanding the crystalline and amorphous solids

Crystalline solids have a well-ordered and repeating three-dimensional arrangement of atoms, ions, or molecules. They have a long-range order, which means that the arrangement of the constituent particles can be predicted over large distances. Amorphous solids, on the other hand, have a disordered atomic arrangement and lack a regular repeating structure. They have only short-range order, which means that the arrangement of the constituent particles can only be predicted over short distances.
02

Explaining X-ray diffraction patterns

X-ray diffraction is a technique used to determine the arrangement of the constituent particles in a solid. When a beam of X-rays is directed onto a solid, the X-ray photons interact with the electron cloud of the atoms, ions or molecules and are scattered in various directions. The scattered X-ray photons interfere constructively or destructively, depending on the distribution and arrangement of the particles in the solid. This interference will produce a diffraction pattern on a detector that can be used to deduce information about the atomic structure of the solid.
03

Compare X-ray diffraction patterns of crystalline and amorphous solids

In the case of crystalline solids, the well-ordered and repeating arrangement of particles leads to a regular interference pattern with discrete and sharp bright spots (known as Bragg peaks), as constructive interference occurs at specific angles. For amorphous solids, the disordered arrangement of particles leads to a diffused interference pattern, with broad and continuous peaks, as the constructive interference occurs over a wider range of angles.
04

Determine which solid produces a simpler X-ray diffraction pattern

Since the X-ray diffraction pattern of a crystalline solid consists of discrete and regular Bragg peaks, it can be considered simpler compared to the diffused, broad, and continuous peaks observed in the diffraction pattern of an amorphous solid. The well-defined peaks in the diffraction pattern of the crystalline solid are due to the long-range order and predictability of the atomic arrangement. Therefore, a crystalline solid will produce a simpler X-ray diffraction pattern compared to an amorphous solid, primarily due to the long-range order of its constituent particles.

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Most popular questions from this chapter

Explain the following: You add \(100 \mathrm{~mL}\) water to a \(500-\mathrm{mL}\) round-bottom flask and heat the water until it is boiling. You remove the heat and stopper the flask, and the boiling stops. You then run cool water over the neck of the flask, and the boiling begins again. It seems as though you are boiling water by cooling it.

A 0.132-mole sample of an unknown semiconducting material with the formula XY has a mass of \(19.0 \mathrm{~g}\). The element \(\mathrm{X}\) 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?

General Zod has sold Lex Luthor what Zod claims to be a new copper-colored form of kryptonite, the only substance that can harm Superman. Lex, not believing in honor among thieves, decided to carry out some tests on the supposed kryptonite. From previous tests, Lex knew that kryptonite is a metal having a specific heat capacity of \(0.082 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and a density of \(9.2 \mathrm{~g} / \mathrm{cm}^{3}\) Lex Luthor's first experiment was an attempt to find the specific heat capacity of kryptonite. He dropped a \(10 \mathrm{~g} \pm 3 \mathrm{~g}\) sample of the metal into a boiling water bath at a temperature of \(100.0^{\circ} \mathrm{C} \pm 0.2^{\circ} \mathrm{C}\). He waited until the metal had reached the bath temperature and then quickly transferred it to \(100 \mathrm{~g} \pm\) \(3 \mathrm{~g}\) of water that was contained in a calorimeter at an initial temperature of \(25.0^{\circ} \mathrm{C} \pm 0.2^{\circ} \mathrm{C}\). The final temperature of the metal and water was \(25.2^{\circ} \mathrm{C}\). Based on these results, is it possible to distinguish between copper and kryptonite? Explain. When Lex found that his results from the first experiment were inconclusive, he decided to determine the density of the sample. He managed to steal a better balance and determined the mass of another portion of the purported kryptonite to be \(4 \mathrm{~g} \pm 1 \mathrm{~g}\). He dropped this sample into water contained in a \(25-\mathrm{mL}\) graduated cylinder and found that it displaced a volume of \(0.42 \mathrm{~mL} \pm 0.02 \mathrm{~mL}\). Is the metal copper or kryptonite? Explain. Lex was finally forced to determine the crystal structure of the metal General Zod had given him. He found that the cubic unit cell contained four atoms and had an edge length of 600\. pm. Explain how this information enabled Lex to identify the metal as copper or kryptonite. Will Lex be going after Superman with the kryptonite or seeking revenge on General Zod? What improvements could he have made in his experimental techniques to avoid performing the crystal structure determination?

The Group 3A/Group \(5 \mathrm{~A}\) semiconductors are composed of equal amounts of atoms from Group \(3 \mathrm{~A}\) and Group \(5 \mathrm{~A}\) - for example, InP and GaAs. These types of semiconductors are used in light- emitting diodes and solid-state lasers. What would you add to make a p-type semiconductor from pure GaAs? How would you dope pure GaAs to make an n-type semiconductor?

An aluminum antimonide solid-state laser emits light with a wavelength of \(730 . \mathrm{nm}\). Calculate the band gap in joules.
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