Textbook Question
Write the electron configurations for the following ions, anddetermine which have noble-gas configurations:(a) Ti2+(b) Br-(c) Mg2+(d) Po2-(e) Pt2+(f) V3+
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Ch.6 - Electronic Structure of Atoms
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232
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Table of contents
- Ch.1 - Introduction: Matter, Energy, and Measurement140
- Ch.2 - Atoms, Molecules, and Ions192
- Ch.3 - Chemical Reactions and Reaction Stoichiometry172
- Ch.4 - Reactions in Aqueous Solution156
- Ch.5 - Thermochemistry151
- Ch.6 - Electronic Structure of Atoms137
- Ch.7 - Periodic Properties of the Elements127
- Ch.8 - Basic Concepts of Chemical Bonding143
- Ch.9 - Molecular Geometry and Bonding Theories167
- Ch.10 - Gases147
- Ch.11 - Liquids and Intermolecular Forces97
- Ch.12 - Solids and Modern Materials129
- Ch.13 - Properties of Solutions126
- Ch.14 - Chemical Kinetics175
- Ch.15 - Chemical Equilibrium107
- Ch.16 - Acid-Base Equilibria127
- Ch.17 - Additional Aspects of Aqueous Equilibria133
- Ch.18 - Chemistry of the Environment88
- Ch.19 - Chemical Thermodynamics140
- Ch.20 - Electrochemistry137
- Ch.21 - Nuclear Chemistry99
- Ch.22 - Chemistry of the Nonmetals66
- Ch.23 - Transition Metals and Coordination Chemistry69
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry11
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Not the one you use?Change textbook
Chapter 6, Problem 43b
One of the emission lines of the hydrogen atom has a wavelength of 94.974 nm. (b) Determine the initial and final values of n associated with this emission.
Verified step by step guidance1
Identify the given wavelength of the emission line, which is 94.974 nm. Convert this wavelength into meters by multiplying by \(10^{-9}\), as 1 nm = \(10^{-9}\) meters.
Use the Rydberg formula for hydrogen emission, which is given by \(\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\), where \(\lambda\) is the wavelength, \(R_H\) is the Rydberg constant (approximately \(1.097 \times 10^7 \, \text{m}^{-1}\)), \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level.
Calculate \(\frac{1}{\lambda}\) using the converted wavelength in meters to find the inverse wavelength in meters\(^{-1}\).
Rearrange the Rydberg formula to solve for \(\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\) by dividing the inverse wavelength by the Rydberg constant.
Estimate possible values of \(n_1\) and \(n_2\) that satisfy the equation \(\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\). Typically, \(n_1\) will be a smaller integer (like 1 or 2) and \(n_2\) will be a larger integer. Check different combinations of \(n_1\) and \(n_2\) to see which pair fits the calculated value from the previous step.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Hydrogen Emission Spectrum
The hydrogen emission spectrum is a series of discrete wavelengths emitted by hydrogen atoms when electrons transition between energy levels. Each line in the spectrum corresponds to a specific transition, with the wavelength related to the energy difference between the initial and final states of the electron.
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Emission Spectra
Energy Level Transitions
In an atom, electrons occupy quantized energy levels, denoted by the principal quantum number n. When an electron moves from a higher energy level (n_initial) to a lower one (n_final), it emits a photon with energy equal to the difference between these levels, which can be calculated using the Rydberg formula.
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Rydberg Formula
The Rydberg formula allows for the calculation of the wavelengths of spectral lines in hydrogen. It is expressed as 1/λ = R_H (1/n_final² - 1/n_initial²), where R_H is the Rydberg constant. This formula is essential for determining the initial and final quantum numbers associated with a given wavelength.
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Related Practice
Textbook Question
One of the emission lines of the hydrogen atom has a wavelength of 94.974 nm. (a) In what region of the electromagnetic spectrum is this emission found?
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Textbook Question
The hydrogen atom can absorb light of wavelength 1094 nm. (a) In what region of the electromagnetic spectrum is this absorption found?
Textbook Question
The Lyman series of emission lines of the hydrogen atom are those for which nf = 1. (a) Determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed.
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Textbook Question
The visible emission lines observed by Balmer all involved nf = 2. (b) Calculate the wavelengths of the first three lines in the Balmer series—those for which ni = 3, 4, and 5—and identify these lines in the emission spectrum shown in Figure 6.11.
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Textbook Question
Order the following transitions in the hydrogen atom from smallest to largest frequency of light absorbed: n = 3 to n = 7, n = 4 to n = 8, n = 2 to n = 5, and n = 1 to n = 3.
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