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Chapter 6: Problem 42

The hydrogen atom can absorb light of wavelength \(2626 \mathrm{nm}\). (a) In what region of the electromagnetic spectrum is this absorption found? (b) Determine the initial and final values of \(n\) associated with this absorption.

Short Answer

Expert verified
The absorption of light with a wavelength of \(2626 nm\) falls in the infrared region of the electromagnetic spectrum. By using the Rydberg formula and trial and error method, we can determine that the initial energy level (n_i) is 1, and the final energy level (n_f) is 2.

Step by step solution

01

(a) Determine the region of the electromagnetic spectrum corresponding to 2626 nm)

To determine the region of the electromagnetic spectrum, we first need to know the wavelength in meters. Given the wavelength of 2626 nm, we can convert it to meters using the following conversion factor: 1 nm = 1 x 10^(-9) m So, 2626 nm = 2626 x 10^(-9) m = 2.626 x 10^(-6) m Now, we can check the electromagnetic spectrum to find the region corresponding to this wavelength. The different regions and their approximate wavelength ranges are: - Radio waves: > 1 x 10^(-1) m - Microwaves: 1 x 10^(-3) - 1 x 10^(-1) m - Infrared: 7 x 10^(-7) - 1 x 10^(-3) m - Visible light: 4 x 10^(-7) - 7 x 10^(-7) m - Ultraviolet: 1 x 10^(-8) - 4 x 10^(-7) m - X-rays: 1 x 10^(-11) - 1 x 10^(-8) m - Gamma-rays: < 1 x 10^(-11) m Looking at these ranges, it can be concluded that the wavelength (2.626 x 10^(-6) m) falls in the infrared region of the electromagnetic spectrum.
02

(b) Determine the initial and final values of n)

To find the initial and final energy levels, we will use the Rydberg formula for the hydrogen atom: \(1/位 = R_H(1/n^2_i - 1/n^2_f)\) Where: - 位 = wavelength (in meters) = 2.626 x 10^(-6) m - R_H = Rydberg constant for hydrogen 鈮� 1.097 x 10^7 m^(-1) - n_i = initial energy level - n_f = final energy level We are given the wavelength and asked to find n_i and n_f values. Since the hydrogen atom is absorbing light energy, the electron must be transitioning from a lower energy level (n_i) to a higher energy level (n_f). First, let's isolate \(n^2_i\) term in the Rydberg formula: \(n^2_i = 1/(位R_H + 1/n^2_f)\) Now we'll use the trial and error method to find the closest integer values of n_i and n_f that satisfy the equation: Let's start with n_f = 2. \(n^2_i = 1/(2.626\cdot10^{-6}\cdot1.097\cdot10^7 + 1/4)\) \(n^2_i 鈮� 1/(4.86 + 0.25)\) \(n^2_i 鈮� 1/5.11 = 0.195\) Now since we know that ni must be an integer value, we can deduce that ni = 1, as it is the closest integer value to 0.195. So, the initial energy level (n_i) is 1. To find the final energy level (n_f), we can use the rearranged Rydberg formula: \(n^2_f = 1/((1/位) - (1/R_H)(1/n^2_i))\) \(n^2_f = 1/((1/(2.626\cdot10^{-6})) - (1/(1.097\cdot10^7))(1/1^2))\) \(n^2_f 鈮� 1/(3.81\cdot10^5 - 1.097\cdot10^7)\) \(n^2_f 鈮� 4.02\) Now, since n_f must also be an integer value, we can deduce that n_f = 2, as it is the closest integer value to 4.02. Therefore, the initial energy level n_i is 1 and the final energy level n_f is 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrogen Atom Energy Levels
The energy levels of an atom correspond to the specific energies that an electron can have within the atom. In the case of the hydrogen atom, which consists of only one electron, these energy levels are quantized. This means that the electron can only exist in certain, discrete energy states. When the electron transitions between these energy levels, it either absorbs or emits energy in the form of light or electromagnetic radiation.

For example, when a hydrogen atom absorbs energy, the electron moves from a lower energy level, denoted as \( n_i \) (initial level), to a higher one, \( n_f \) (final level). These energy levels are labeled with quantum numbers (n), where \( n = 1 \) is the ground state (lowest energy level), and \( n = 2, 3, 4,\ldots\) are excited states (higher energy levels).

  • At \( n = 1 \), the electron is closest to the nucleus and has the lowest energy.
  • As the quantum number increases (\( n = 2, 3,\ldots\)), the energy level increases, meaning the electron is further from the nucleus and has higher energy.
Understanding these energy levels is crucial for explaining how hydrogen atoms emit or absorb specific wavelengths of light, as described by the Rydberg formula.
Rydberg Formula
The Rydberg formula is a mathematical equation used to predict the wavelengths of light emitted or absorbed by electrons in a hydrogen atom. It helps us determine the possible transitions between different energy levels of the electron. This formula is especially useful because it provides a quantitative relationship between the wavelengths of light and the quantum numbers of the initial and final energy levels.

The Rydberg formula is written as:\[\frac{1}{\lambda} = R_H\left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right)\]Where:
  • \( \lambda \) is the wavelength of the absorbed or emitted light.
  • \( R_H \) is the Rydberg constant for hydrogen, approximately \( 1.097 \times 10^7 \) m\(^-1\).
  • \( n_i \) and \( n_f \) are the initial and final quantum numbers of the energy levels.
When a hydrogen electron absorbs light, it moves from a lower energy level (\( n_i \)) to a higher one (\( n_f \)). Conversely, when emitting light, the transition is from a higher to a lower energy level. By substituting known wavelengths into the Rydberg formula, we can calculate the corresponding values of \( n_i \) and \( n_f \), helping us understand the nature of the electronic transitions.
Infrared Region
The infrared region of the electromagnetic spectrum includes wavelengths longer than visible light but shorter than microwaves, roughly spanning from \( 700 \) nm to \( 1 \) mm. Infrared light is generally experienced as heat and is not visible to the human eye.

In the context of hydrogen atom transitions, wavelengths that fall within the infrared range often result from electrons moving between lower energy levels. This is because transitions involving larger energy changes typically produce wavelengths in the visible or ultraviolet regions, while those involving smaller energy changes result in infrared light.

For example, if the wavelength is \( 2626 \) nm, as in the original exercise, it clearly falls in the infrared region. Here, the energy transition is enough to move the electron between nearby energy levels, such as from \( n_i = 1 \) to \( n_f = 2 \), without requiring the higher energy typical of ultraviolet or visible light transitions.

Understanding the infrared region's characteristics helps in identifying the type of electronic transitions occurring in an atom and is essential in spectroscopy and other fields that involve analyzing light interactions.

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Most popular questions from this chapter

The discovery of hafnium, element number \(72,\) provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements \(58-71\) ) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), is reacted with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and CO in the ratio 1: 2 . Write a balanced chemical equation for the reaction. Starting with a 55.4-g sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and \(\mathrm{Hf}\) form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\)

Certain elements emit light of a specific wavelength when they are burned. Historically, chemists used such emission wavelengths to determine whether specific elements were present in a sample. Characteristic wavelengths for some of the elements are given in the following table: \(\begin{array}{llll}\mathrm{Ag} & 328.1 \mathrm{nm} & \mathrm{Fe} & 372.0 \mathrm{nm} \\ \mathrm{Au} & 267.6 \mathrm{nm} & \mathrm{K} & 404.7 \mathrm{nm} \\ \mathrm{Ba} & 455.4 \mathrm{nm} & \mathrm{Mg} & 285.2 \mathrm{nm} \\ \mathrm{Ca} & 422.7 \mathrm{nm} & \mathrm{Na} & 589.6 \mathrm{nm} \\ \mathrm{Cu} & 324.8 \mathrm{nm} & \mathrm{Ni} & 341.5 \mathrm{nm}\end{array}\) (a) Determine which elements emit radiation in the visible part of the spectrum. (b) Which element emits photons of highest energy? Of lowest energy? (c) When burned, a sample of an unknown substance is found to emit light of frequency \(6.59 \times 10^{14} \mathrm{~s}^{-1} .\) Which of these elements is probably in the sample?

(a) What is the frequency of radiation whose wavelength is \(5.0 \times 10^{-5} \mathrm{~m} ?\) (b) What is the wavelength of radiation that has a frequency of \(2.5 \times 10^{8} \mathrm{~s}^{-1} ?(\mathrm{c})\) Would the radiations in part (a) or part (b) be detected by an X-ray detector? (d) What distance does electromagnetic radiation travel in \(10.5 \mathrm{fs}\) ?

It is possible to convert radiant energy into electrical energy using photovoltaic cells. Assuming equal efficiency of conversion, would infrared or ultraviolet radiation yield more electrical energy on a per-photon basis?

(a) What are the similarities and differences between the \(1 s\) and \(2 s\) orbitals of the hydrogen atom? (b) In what sense does a \(2 p\) orbital have directional character? Compare the "directional" characteristics of the \(p_{x}\) and \(d_{x^{2}-y^{2}}\) orbitals. (That is, in what direction or region of space is the electron density concentrated?) (c) What can you say about the average distance from the nucleus of an electron in a \(2 s\) orbital as compared with a \(3 s\) orbital? (d) For the hydrogen atom, list the following orbitals in order of increasing energy (that is, most stable ones first): \(4 f, 6 s, 3 d, 1 s, 2 p\).
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