Chapter 6: Problem 54
Excited \(\mathrm{H}\) atoms have many emission lines. One series of lines, called the \(Pfund series\), occurs in the infrared region. It results when an electron changes from higher energy levels to a level with \(n=5 .\) Calculate the wavelength and frequency of the lowest energy line of this series.
Short Answer
Expert verified
The wavelength is \( 7.46 \times 10^{-6} \, \mathrm{m} \) and the frequency is \( 4.02 \times 10^{13} \, \mathrm{Hz} \).
Step by step solution
01
Understand the Pfund series
The Pfund series occurs when electron transitions end at the energy level \( n=5 \). The lowest energy line in this series corresponds to the transition from \( n=6 \) to \( n=5 \).
02
Rydberg formula application
Use the Rydberg formula to find the wavelength \( \lambda \) of the line: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), \]where \( n_1 = 5 \), \( n_2 = 6 \), and \( R = 1.097 \times 10^7 \, \mathrm{m^{-1}} \) is the Rydberg constant.
03
Calculate the wavelength
Substitute the values into the Rydberg formula:\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{5^2} - \frac{1}{6^2} \right) \]\[ = 1.097 \times 10^7 \left( \frac{1}{25} - \frac{1}{36} \right) \]\[ = 1.097 \times 10^7 \left( 0.04 - 0.02777778 \right) \]\[ = 1.097 \times 10^7 \times 0.01222222 \]\[ = 1.34 \times 10^5 \mathrm{m^{-1}} \].The wavelength \( \lambda \) is thus \[ \lambda = \frac{1}{1.34 \times 10^5} = 7.46 \times 10^{-6} \, \mathrm{m} \].
04
Calculate the frequency
Use the speed of light equation to find the frequency \( u \):\[ c = \lambda u \]where \( c = 3 \times 10^8 \, \mathrm{m/s} \). Solve for frequency:\[ u = \frac{c}{\lambda} = \frac{3 \times 10^8}{7.46 \times 10^{-6}} \]\[ = 4.02 \times 10^{13} \, \mathrm{Hz} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Pfund Series
The Pfund series is an interesting group of spectral lines in atomic physics. It involves the emission of light when electrons in a hydrogen atom transition from higher energy levels back down to a specific energy state: the level with a principal quantum number of \(n=5\).
This series is distinguished by its position in the infrared region of the electromagnetic spectrum.
This means that the wavelengths of the Pfund series are too long to be visible to the human eye, making them detectable only with specialized equipment designed for infrared wavelengths.
This series is distinguished by its position in the infrared region of the electromagnetic spectrum.
This means that the wavelengths of the Pfund series are too long to be visible to the human eye, making them detectable only with specialized equipment designed for infrared wavelengths.
- Origin of the Name: The series is named after the American physicist August Herman Pfund, who first discovered it in 1924.
- Historic Importance: The Pfund series was essential in providing evidence of quantized energy levels within the atom, a critical validation of the Bohr model of the atom.
Electron Transition
Electron transition refers to the movement of an electron between different orbits or energy levels within an atom. These transitions happen when an electron absorbs or releases energy, often in the form of light.
In the context of the Pfund series, we focus on transitions that end at the fifth energy level.
The lowest energy line of this series involves an electron transition from \(n=6\) to \(n=5\).
In the context of the Pfund series, we focus on transitions that end at the fifth energy level.
The lowest energy line of this series involves an electron transition from \(n=6\) to \(n=5\).
- Energy Change: When an electron loses energy, it moves to a lower energy level, emitting a photon whose energy matches the difference in energy between the two levels.
- Significance of Lower Levels: Lower energy levels, like \(n=5\), indicate the end point of the transition and determine the series to which the spectral line belongs.
- Balmer Series Comparison: By contrast, in the Balmer series, electrons transition to the \(n=2\) energy level, producing visible light.
Spectral Lines
Spectral lines are fundamental in identifying and understanding elements. They appear when light is emitted or absorbed by electrons as they transition between different energy levels. These transitions create lines at specific wavelengths, seen as dark absorption lines or bright emission lines on a spectrum.
The Pfund series spectral lines are part of emission spectra and occur in the infrared region.
The Pfund series spectral lines are part of emission spectra and occur in the infrared region.
- Types of Spectra: Emission spectra result from electrons falling to lower energy states, emitting light, whereas absorption spectra occur when electrons absorb light to jump to higher levels.
- Fingerprints of Elements: Each element has unique spectral lines, acting like a fingerprint that helps scientists identify elements in stars and other celestial bodies.
- Applications: Spectral lines are used in astronomy for determining the chemical composition of stars and in laboratories for identifying substances.
Infrared Wavelength Calculation
Calculating the wavelength of infrared emissions in spectral lines involves the Rydberg formula, which is essential for understanding energy transitions. In the Pfund series, we utilize the Rydberg formula to derive the wavelength for the lowest energy spectral line when the change occurs from \(n=6\) to \(n=5\).
The formula is expressed as: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \(R\) is the Rydberg constant approximately equal to \(1.097 \times 10^7 \, \mathrm{m^{-1}}\).
The formula is expressed as: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \(R\) is the Rydberg constant approximately equal to \(1.097 \times 10^7 \, \mathrm{m^{-1}}\).
- Application of Formula: Substitute \(n_1=5\) and \(n_2=6\) into the formula to find the wavelength \(\lambda\).
- Result: Performing this calculation gives a wavelength in the infrared spectrum, about \(7.46 \times 10^{-6} \, \mathrm{m}\).
- Frequency Calculation: Once the wavelength is known, the speed of light formula \(c = \lambda \times u\) allows you to determine the frequency \(u\), calculated as approximately \(4.02 \times 10^{13} \, \mathrm{Hz}\).