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The Lyman series of emission lines of the hydrogen atom are those for which \(n_{\mathrm{f}}=1 .\) (a) Determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed. (b) Calculate the wavelengths of the first three lines in the Lyman series-those for which \(n_{\mathrm{i}}=2,3,\) and 4 .

Short Answer

Expert verified
The Lyman series is in the ultraviolet region. The first three lines are at 121.5 nm, 102.8 nm, and 97.2 nm.

Step by step solution

01

Understand the Lyman Series

The Lyman series refers to electronic transitions in the hydrogen atom where electrons fall from higher energy levels to the first energy level (\(n_f = 1\)). This series results in emission lines in the ultraviolet region of the electromagnetic spectrum.
02

Use Rydberg Formula for Wavelengths

The Rydberg formula for the wavelengths of hydrogen spectral lines is given by:\[\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\]where \(R_H\) is the Rydberg constant, \(1.097 \times 10^7 \text{ m}^{-1}\). For the Lyman series, \(n_f = 1\), and \(n_i\) is the initial energy level.
03

Calculate Wavelength for \(n_i = 2\)

Substitute \(n_f = 1\) and \(n_i = 2\) into the Rydberg formula:\[\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)\]\[\frac{1}{\lambda} = 1.097 \times 10^7 \left( 1 - \frac{1}{4} \right) = 1.097 \times 10^7 \times \frac{3}{4} = 8.228 \times 10^6 \text{ m}^{-1}\]\[\lambda = \frac{1}{8.228 \times 10^6} \approx 1.215 \times 10^{-7} \text{ m} = 121.5 \text{ nm}\]
04

Calculate Wavelength for \(n_i = 3\)

Substitute \(n_f = 1\) and \(n_i = 3\) into the Rydberg formula:\[\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{3^2} \right)\]\[\frac{1}{\lambda} = 1.097 \times 10^7 \left( 1 - \frac{1}{9} \right) = 1.097 \times 10^7 \times \frac{8}{9} = 9.752 \times 10^6 \text{ m}^{-1}\]\[\lambda = \frac{1}{9.752 \times 10^6} \approx 1.028 \times 10^{-7} \text{ m} = 102.8 \text{ nm}\]
05

Calculate Wavelength for \(n_i = 4\)

Substitute \(n_f = 1\) and \(n_i = 4\) into the Rydberg formula:\[\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{4^2} \right)\]\[\frac{1}{\lambda} = 1.097 \times 10^7 \left( 1 - \frac{1}{16} \right) = 1.097 \times 10^7 \times \frac{15}{16} = 1.029 \times 10^7 \text{ m}^{-1}\]\[\lambda = \frac{1}{1.029 \times 10^7} \approx 9.72 \times 10^{-8} \text{ m} = 97.2 \text{ nm}\]

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Key Concepts

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

Hydrogen Atom
The hydrogen atom is the simplest type of atom, consisting of just one proton and one electron. It serves as an excellent model for studying atomic structure due to its simplicity and well-understood properties. In the hydrogen atom, the electron is bound to the proton by electromagnetic forces, and it can only occupy specific energy levels, known as quantized energy levels.
When the electron jumps from one energy level to another, it absorbs or emits energy in the form of light, producing spectral lines. These transitions are key to understanding how atoms emit and absorb light.
For the hydrogen atom, these transitions lead to various spectral series, such as the Lyman series in the ultraviolet spectrum, providing insight into the energy architecture of the atom. Understanding hydrogen atoms allows us to explore not just chemistry and physics, but also aids in comprehending more complex atomic interactions.
Rydberg Formula
The Rydberg formula is a mathematical equation used to predict the wavelengths of spectral lines of hydrogen. It is fundamental in the study of atomic spectra and helps explain the spectral line emissions that occur when an electron transitions between energy levels.
The formula is expressed as follows:\[\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\]where:
  • \(\lambda\) is the wavelength of the emitted light.
  • \(R_H\) is the Rydberg constant, approximately \(1.097 \times 10^7 \text{ m}^{-1}\).
  • \(n_f\) is the final energy level.
  • \(n_i\) is the initial energy level the electron descends from.

For the Lyman series in the hydrogen atom, \(n_f\) is always 1, indicating that the electron transitions to the ground state, releasing energy as ultraviolet light. The formula's predictive power provides a way to calculate the precise wavelengths of these light emissions, enhancing our understanding of atomic transitions.
Ultraviolet Spectrum
The ultraviolet (UV) spectrum consists of electromagnetic waves with wavelengths shorter than visible light but longer than X-rays, typically in the range of about 10 nm to 400 nm. UV light is invisible to the human eye, though it can cause fluorescence in certain substances, making them appear to glow when exposed.
The Lyman series of the hydrogen atom falls into this part of the spectrum due to the high energy nature of the transitions when electrons return to the first energy level from higher levels. This shows how the energy levels in an atom determine the nature of light but also the electromagnetic spectrum segment in which the light is emitted.
Studying these emissions provides insight into how electromagnetic spectra are connected to atomic interactions, and it is valuable in fields such as astronomy and molecular biology, where understanding the behavior of light and matter at this scale is critical.
Energy Levels
Energy levels in an atom define the specific energies that electrons can have when bound to an atom. In the hydrogen atom, these energy levels are quantized, meaning that they can only take on particular values and no others.
The energies of these levels are determined by the principle quantum number \(n\), which is an integer. The lower the value of \(n\), the lower the energy and closer the electron is to the nucleus. Transitions of electrons between these energy levels release or absorb specific amounts of energy, leading to the emission or absorption of light at particular wavelengths.
In the case of the Lyman series, electrons drop to the first energy level from higher levels, leading to the emission of ultraviolet light. Each possible transition corresponds to a different spectral line. These quantized energy levels explain the distinct spectral lines seen in the electromagnetic spectrum, illustrating the principles of quantum mechanics in atomic physics.

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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, \(\mathrm{ZrO}_{2}\), reacts with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) 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 \(\mathrm{MO}_{2}\).

A diode laser emits at a wavelength of \(987 \mathrm{nm} .\) (a) In what portion of the electromagnetic spectrum is this radiation found? (b) All of its output energy is absorbed in a detector that measures a total energy of \(0.52 \mathrm{~J}\) over a period of \(32 \mathrm{~s}\). How many photons per second are being emitted by the laser?

The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of \(9.47 \times 10^{6} \mathrm{~m} / \mathrm{s}\) What is the characteristic wavelength of this electron? Is the wavelength comparable to the size of atoms?

Determine which of the following statements are false and correct them. (a) The frequency of radiation increases as the wavelength increases. (b) Electromagnetic radiation travels through a vacuum at a constant speed, regardless of wavelength. (c) Infrared light has higher frequencies than visible light. (d) The glow from a fireplace, the energy within a microwave oven, and a foghorn blast are all forms of electromagnetic radiation.

Identify the group of elements that corresponds to each of the following generalized electron configurations and indicate the number of unpaired electrons for each: (a) [noble gas]ns \({ }^{2} n p^{5}\) (b) [noble gas] \(n s^{2}(n-1) d^{2}\) (c) \([\) noble \(\operatorname{gas}] n s^{2}(n-1) d^{10} n p^{1}\) (d) [noble gas \(] n s^{2}(n-2) f^{6}\)

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