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Chapter 2: Problem 34

Scientists use emission spectra to confirm the presence of an element in materials of unknown composition. Why is this possible?

Short Answer

Expert verified
Scientists use emission spectra to confirm the presence of an element in materials of unknown composition because each element has a unique set of energy levels for its electrons, which results in a unique emission spectrum when its electrons are excited and then return to lower energy levels, releasing energy as light. By comparing a sample's emission spectrum to known emission spectra of various elements, scientists can identify and confirm the presence of specific elements in the material, making it a reliable method for analyzing unknown compositions.

Step by step solution

01

Understand Emission Spectra

Emission spectra are the unique patterns of wavelengths (colors) of light emitted by an energy source, such as a heated or irradiated substance. Each element has a unique set of energy levels for its electrons, and when electrons in these elements are excited (given additional energy), they jump to higher energy levels. However, the electrons aren't stable at these higher energy levels, and they return to the original, lower energy levels. In the process, they release energy in the form of light, which generates the characteristic emission spectrum for the element.
02

Generate Emission Spectra

To generate an emission spectrum from a sample, a scientist will first expose the sample to a high-energy source, such as an electric discharge or a flame. This energy source excites the electrons in the atoms of the sample, causing them to jump to higher energy levels. When the electrons return to their original energy levels, they emit light that can be analyzed with a spectroscope, which separates the light into its component wavelengths. This produces a pattern of bright lines, called the emission spectrum, unique to the element being analyzed.
03

Identify Elements by Unique Spectra

Since each element has a unique set of energy levels for its electrons, the emitted light - and thus, the emission spectrum - is characteristic of the element. This means that when a sample's emission spectrum is observed, it can be compared to known emission spectra of various elements. If the emission spectrum of the sample matches the known emission spectrum of a particular element, then that confirms the presence of the element in the material being analyzed.
04

Application to Unknown Composition Analysis

When analyzing materials of unknown composition, emission spectra can be used to determine which elements are present in the sample as well as their relative concentrations. By comparing the emission spectra of the sample to the known emission spectra of various elements, scientists can identify and confirm the presence of specific elements in the material. This method is widely used in areas such as environmental testing, forensic science, and quality control in manufacturing, to ensure the materials being used meet required specifications.

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Most popular questions from this chapter

Valence electrons are those electrons in the outermost principal quantum level (highest \(n\) level) of an atom in its ground state. Groups \(1 \mathrm{A}\) to \(8 \mathrm{A}\) have from 1 to 8 valence electrons. For each group of the representative elements (1A-8A), give the number of valence electrons, the general valence electron configuration, a sample element in that group, and the specific valence electron configuration for that element.

Which of elements \(1-36\) have two unpaired electrons in the ground state?

Ionization energy is the energy required to remove an electron from an atom in the gas phase. The ionization energy of gold is \(890.1 \mathrm{kJ} / \mathrm{mol} .\) Is light with a wavelength of \(225 \mathrm{nm}\) capable of ionizing a gold atom (removing an electron) in the gas phase? ( 1 mol gold \(=6.022 \times 10^{23}\) atoms gold.)

Which of the following electron configurations correspond to an excited state? Identify the atoms and write the ground-state electron configuration where appropriate. a. \(1 s^{2} 2 s^{2} 3 p^{1}\) b. \(1 s^{2} 2 s^{2} 2 p^{6}\) c. \(1 s^{2} 2 s^{2} 2 p^{4} 3 s^{1}\) d. \([\mathrm{Ar}] 4 s^{2} 3 d^{5} 4 p^{1}\) How many unpaired electrons are present in each of these species?

Consider the following ionization energies for aluminum: $$ \begin{aligned} \mathrm{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} & I_{1}=580 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} & I_{2}=1815 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} & I_{3}=2740 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} & I_{4}=11,600 \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\).
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