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Ch.6 - Electronic Structure of Atoms
Chapter 6, Problem 42b

The Lyman series of emission lines of the hydrogen atom are those for which nf = 1. (b) Calculate the wavelengths of the first three lines in the Lyman series—those for which ni = 2, 3, and 4.

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Identify the formula to calculate the wavelength of the emission lines in the hydrogen spectrum, which is given by the Rydberg formula: \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \), where \( \lambda \) is the wavelength, \( R_H \) is the Rydberg constant (approximately 1.097 x 10^7 m^-1), \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.
Set \( n_f = 1 \) for the Lyman series as given in the problem statement.
Calculate the wavelength for the first line in the series where \( n_i = 2 \). Plug these values into the Rydberg formula and solve for \( \lambda \).
Repeat the calculation for the second line in the series where \( n_i = 3 \).
Repeat the calculation for the third line in the series where \( n_i = 4 \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lyman Series

The Lyman series refers to a set of spectral lines corresponding to transitions of electrons in a hydrogen atom from higher energy levels (ni) to the lowest energy level (nf = 1). These transitions result in the emission of ultraviolet light, and the wavelengths of these lines can be calculated using the Rydberg formula.
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Rydberg Formula

The Rydberg formula is an equation used to predict the wavelengths of spectral lines in hydrogen and other hydrogen-like atoms. It is expressed as 1/λ = R_H (1/nf^2 - 1/ni^2), where R_H is the Rydberg constant, λ is the wavelength, nf is the final energy level, and ni is the initial energy level. This formula is essential for calculating the wavelengths of the Lyman series.
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Energy Levels in Hydrogen Atom

In a hydrogen atom, electrons occupy discrete energy levels, which are quantized. The energy levels are defined by the principal quantum number n, where n = 1 is the ground state. When an electron transitions between these levels, energy is emitted or absorbed in the form of photons, with the energy difference corresponding to the wavelength of the emitted light.
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Related Practice
Textbook Question

The visible emission lines observed by Balmer all involved nf = 2. (a) Which of the following is the best explanation of why the lines with nf = 3 are not observed in the visible portion of the spectrum: (i) Transitions to nf = 3 are not allowed to happen, (ii) transitions to nf = 3 emit photons in the infrared portion of the spectrum, (iii) transitions to nf = 3 emit photons in the ultraviolet portion of the spectrum, or (iv) transitions to nf = 3 emit photons that are at exactly the same wavelengths as those to nf = 2.

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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

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

One of the emission lines of the hydrogen atom has a wavelength of 93.07 nm. a. In what region of the electromagnetic spectrum is this emission found?

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Textbook Question

One of the emission lines of the hydrogen atom has a wavelength of 93.07 nm. b. Determine the initial and final values of n associated with this emission.

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Textbook Question

The hydrogen atom can absorb light of wavelength 1094 nm. (b) Determine the final value of n associated with this absorption.

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