EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
7th Edition
ISBN: 9780100257047
Author: Chang
Publisher: YUZU
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Chapter 7, Problem 7.106QP
Interpretation Introduction

Interpretation:

The wavelengths in the increasing order of the first four transitions in the Balmer series of the He+ ion, comparing these wavelengths with the same transitions in an H atom and the comment on the differences between these wavelengths of He+ ion and H atom which occur in the electromagnetic radiation should be described by applying the Rydberg equation in them.

Concept Introduction:

Absorption refers to how much light can be taken in by the material being measured.

Emission spectrum:

When electromagnetic radiation interacts with matter, atoms and molecules may absorb energy and reach to a higher energy state.  With higher energy, these are in an unstable state.  For returning to their normal (more stable, lower energy) energy state, the atoms and molecules emit radiations in various regions of the electromagnetic spectrum.  The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum.

In 1885, Johann Balmer developed a simple equation which could be used to calculate the wavelengths of the four visible lines in the emission spectrum of hydrogen.  Johannes Rydberg developed Balmer’s equation further, giving an equation which could calculate the visible wavelengths and also those of all hydrogen’s spectral lines.

1λ = R(1n121n22)

This equation is known as the Rydberg equation.  Here, λ is the wavelength of a line in the spectrum; R is the Rydberg constant (1.09737316 × 107 m1 for H atom); and n1 and n2 are positive integers, where n2 > n1.

Expert Solution & Answer
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Answer to Problem 7.106QP

The wavelengths of the first four transitions in the Balmer series of the He+ ion are 164 nm, 121 nm, 108 nm and 103 nm whereas for H atom, they are 656 nm, 486 nm, 434 nm and 410 nm.  All the Balmer transitions for He+ are in the ultraviolet region whereas the transitions for H atom are all in the visible region.

Explanation of Solution

When one of helium’s electrons is removed, the resulting species is the helium ion, He+.  The He+ ion contains only one electron and is therefore a “hydrogen-like ion”.  Balmer series corresponds to transitions to = 2 level.  The first four transitions in the Balmer series mean the transitions from = 3, 4, 5 and 6 energy levels to the transition = 2 energy level.  The formula to calculate the wavelength from Rydberg equation is

λ = 1R(1n121n22)

Here, the Rydberg constant for He is 4.39 × 107 m1.

For the transition = 3  n = 2, the wavelength of the He+ ion is calculated as follows:

λ = 1R(1n121n22)λ = 14.39 × 107m1 (132122)λ = 1.64 × 107 m

The negative sign indicates that the emission of light occurs.  Wavelengths are always positive signs.  Here, 1 nm = 109 m

λ = 1.64 × 10mλ = 1.64 × 10m × 10nm1 mλ = 164 nm

For the transition = 3  n = 2, the wavelength of the He+ ion is 164 nm.  Simiarly, for the transition = 4  n = 2, the wavelength of the He+ ion is

λ = 14.39 × 107 m1(142122)λ = 121 nm

For the transition = 5  n = 2, the wavelength of the He+ ion is

λ = 14.39 × 107 m1(152122)λ = 108 nm

For the transition = 6  n = 2, the wavelength of the He+ ion is

λ = 14.39 × 107 m1(162122)λ = 103 nm

The Rydberg constant for H atom is 1.097 × 107 m1.  For the transition = 3  n = 2, the wavelength of the H atom is calculated as follows:

λ = 1R(1n121n22)λ = 11.097 × 107 m1(132122)λ = 6.56 × 107 m

The negative sign indicates that the emission of light occurs.  Wavelengths are always positive signs.  Here, 1 nm = 109 m

λ = 6.56 × 107 mλ = 6.56 × 107m × 109 nm1 mλ = 656 nm

For the transition = 3  n = 2, the wavelength of the H atom is 656 nm.  Simiarly, for the transition = 4  n = 2, the wavelength of the H atom is

λ = 11.097 × 107 m1(142122)λ = 486 nm

For the transition = 5  n = 2, the wavelength of the H atom is

λ = 11.097 × 107 m1 (152122)λ = 434 nm

For the transition = 6  n = 2, the wavelength of the H atom is

λ = 11.097 × 107 m1 (162122)λ = 434 nm

All the Balmer transitions for He+ are in the ultraviolet region whereas the transitions for H atom are all in the visible region.

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Chapter 7 Solutions

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO

Ch. 7.5 - Prob. 1RCCh. 7.6 - Prob. 1RCCh. 7.7 - Prob. 1PECh. 7.7 - Prob. 2PECh. 7.7 - Prob. 1RCCh. 7.8 - Prob. 1PECh. 7.8 - Prob. 2PECh. 7.8 - Prob. 3PECh. 7.8 - Prob. 1RCCh. 7.9 - Prob. 1PECh. 7.9 - Prob. 1RCCh. 7 - Prob. 7.1QPCh. 7 - Prob. 7.2QPCh. 7 - Prob. 7.3QPCh. 7 - Prob. 7.4QPCh. 7 - Prob. 7.5QPCh. 7 - Prob. 7.6QPCh. 7 - Prob. 7.7QPCh. 7 - 7.8 (a) What is the frequency of tight having a...Ch. 7 - Prob. 7.9QPCh. 7 - Prob. 7.10QPCh. 7 - Prob. 7.11QPCh. 7 - 7.12 The SI unit of length is the meter, which...Ch. 7 - 7.13 What are photons? What role did Einstein's...Ch. 7 - Prob. 7.14QPCh. 7 - Prob. 7.15QPCh. 7 - Prob. 7.16QPCh. 7 - Prob. 7.17QPCh. 7 - Prob. 7.18QPCh. 7 - Prob. 7.19QPCh. 7 - Prob. 7.20QPCh. 7 - Prob. 7.21QPCh. 7 - Prob. 7.22QPCh. 7 - Prob. 7.23QPCh. 7 - Prob. 7.24QPCh. 7 - Prob. 7.25QPCh. 7 - Prob. 7.26QPCh. 7 - Prob. 7.27QPCh. 7 - Prob. 7.28QPCh. 7 - Prob. 7.29QPCh. 7 - Prob. 7.30QPCh. 7 - Prob. 7.31QPCh. 7 - Prob. 7.32QPCh. 7 - Prob. 7.33QPCh. 7 - Prob. 7.34QPCh. 7 - Prob. 7.35QPCh. 7 - Prob. 7.36QPCh. 7 - Prob. 7.37QPCh. 7 - Prob. 7.38QPCh. 7 - Prob. 7.39QPCh. 7 - Prob. 7.40QPCh. 7 - Prob. 7.41QPCh. 7 - 7.42 What is the de Broglie wavelength (in nm)...Ch. 7 - Prob. 7.43QPCh. 7 - Prob. 7.44QPCh. 7 - Prob. 7.45QPCh. 7 - Prob. 7.46QPCh. 7 - Prob. 7.47QPCh. 7 - Prob. 7.48QPCh. 7 - 7.49 Why is a boundary surface diagram useful in...Ch. 7 - Prob. 7.50QPCh. 7 - Prob. 7.51QPCh. 7 - Prob. 7.52QPCh. 7 - Prob. 7.53QPCh. 7 - Prob. 7.54QPCh. 7 - Prob. 7.55QPCh. 7 - Prob. 7.56QPCh. 7 - Prob. 7.57QPCh. 7 - 7.58 What is the difference between a 2px and a...Ch. 7 - Prob. 7.59QPCh. 7 - Prob. 7.60QPCh. 7 - Prob. 7.61QPCh. 7 - Prob. 7.62QPCh. 7 - Prob. 7.63QPCh. 7 - Prob. 7.64QPCh. 7 - 7.65 Make a chart of all allowable orbitals in the...Ch. 7 - 7.66 Why do the 3s, 3p, and 3d orbitals have the...Ch. 7 - Prob. 7.67QPCh. 7 - Prob. 7.68QPCh. 7 - Prob. 7.69QPCh. 7 - Prob. 7.70QPCh. 7 - Prob. 7.71QPCh. 7 - Prob. 7.72QPCh. 7 - Prob. 7.73QPCh. 7 - Prob. 7.74QPCh. 7 - Prob. 7.75QPCh. 7 - Prob. 7.76QPCh. 7 - Prob. 7.77QPCh. 7 - 7.78 Comment on the correctness of the following...Ch. 7 - Prob. 7.79QPCh. 7 - Prob. 7.80QPCh. 7 - Prob. 7.81QPCh. 7 - Prob. 7.82QPCh. 7 - Prob. 7.83QPCh. 7 - Prob. 7.84QPCh. 7 - Prob. 7.85QPCh. 7 - Prob. 7.86QPCh. 7 - Prob. 7.87QPCh. 7 - Prob. 7.88QPCh. 7 - Prob. 7.89QPCh. 7 - Prob. 7.90QPCh. 7 - Prob. 7.91QPCh. 7 - Prob. 7.92QPCh. 7 - Prob. 7.93QPCh. 7 - Prob. 7.94QPCh. 7 - 7.95 Identify the following individuals and their...Ch. 7 - Prob. 7.96QPCh. 7 - Prob. 7.97QPCh. 7 - Prob. 7.98QPCh. 7 - Prob. 7.99QPCh. 7 - 7.100 A laser is used in treating retina...Ch. 7 - 7.101 A 368-g sample of water absorbs infrared...Ch. 7 - Prob. 7.102QPCh. 7 - Prob. 7.103QPCh. 7 - Prob. 7.104QPCh. 7 - Prob. 7.105QPCh. 7 - Prob. 7.106QPCh. 7 - Prob. 7.107QPCh. 7 - Prob. 7.108QPCh. 7 - Prob. 7.109QPCh. 7 - Prob. 7.110QPCh. 7 - Prob. 7.111QPCh. 7 - 7.112 An atom moving at its root-mean-square speed...Ch. 7 - Prob. 7.113QPCh. 7 - Prob. 7.114QPCh. 7 - Prob. 7.115QPCh. 7 - Prob. 7.116QPCh. 7 - Prob. 7.117SPCh. 7 - Prob. 7.118SPCh. 7 - Prob. 7.119SPCh. 7 - Prob. 7.120SPCh. 7 - 7.121 According to Einstein’s special theory of...Ch. 7 - Prob. 7.122SPCh. 7 - Prob. 7.123SPCh. 7 - Prob. 7.124SPCh. 7 - Prob. 7.125SPCh. 7 - 7.126 The wave function for the 2s orbital in the...Ch. 7 - Prob. 7.127SPCh. 7 - Prob. 7.128SPCh. 7 - Prob. 7.129SP
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