/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Calculate the wavelengths of the... [FREE SOLUTION] | 91影视

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Calculate the wavelengths of the first three lines of the Lyman, Balmer, and Paschen series, and the series limit (the shortest wavelength) for each series.

Short Answer

Expert verified
The first three wavelengths of the Lyman series are found using the Rydberg formula with \( n_1 = 1 \) and \( n_2 = 2, 3, 4 \), yielding \(\lambda_{L1}, \lambda_{L2}, \lambda_{L3}\). Similarly, for the Balmer series with \( n_1 = 2 \) and \(n_2 = 3, 4, 5\), the wavelengths are \(\lambda_{B1}, \lambda_{B2}, \lambda_{B3}\), and for the Paschen series with \( n_1 = 3 \) and \( n_2 = 4, 5, 6 \), the wavelengths are \(\lambda_{P1}, \lambda_{P2}, \lambda_{P3}\). The series limits are found by setting \( n_2 \) to infinity in the Rydberg formula, obtaining the shortest wavelengths for Lyman, Balmer, and Paschen series as \(\lambda_{L_{lim}}, \lambda_{B_{lim}}, \lambda_{P_{lim}}\).

Step by step solution

01

Calculate wavelengths for the Lyman series

The Lyman series corresponds to transitions where the final energy level \( n_1 = 1 \). The first three lines of the Lyman series are transitions from \( n_2 = 2, 3, 4 \). Applying the Rydberg formula: - For \( n_2 = 2 \): \[ 1/\lambda = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 1.097373 x 10^7 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) m^{-1} \] Solving for \( \lambda \) gives the wavelength of the first line. - Similarly, for \( n_2 = 3 \) and \( n_2 = 4 \), one can find the wavelengths of the second and third lines.
02

Calculate wavelengths for the Balmer series

The Balmer series corresponds to transitions where the final energy level \( n_1 = 2 \). The first three lines are transitions from \( n_2 = 3, 4, 5 \). By substituting these values into the Rydberg formula, one can obtain the corresponding wavelengths.
03

Calculate wavelengths for the Paschen series

The Paschen series corresponds to transitions where the final energy level \( n_1 = 3 \). The first three lines are transitions from \( n_2 = 4, 5, 6 \). By substituting these values into the Rydberg formula, one can obtain the corresponding wavelengths.
04

Calculate series limits for each series

The series limit for each series is found by setting \( n_2 \) to infinity in the Rydberg formula. This is the shortest possible wavelength for each series, and corresponds to the greatest amount of energy that can be absorbed or emitted: - For the Lyman series: \( 1/\lambda = 1.097373 x 10^7 m^{-1} / 1^2 \) - For the Balmer series: \( 1/\lambda = 1.097373 x 10^7 m^{-1} / 2^2 \) - For the Paschen series: \( 1/\lambda = 1.097373 x 10^7 m^{-1} / 3^2 \) Using these formulas, the series limits for each series can be found by solving for \( \lambda \).

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Key Concepts

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

Lyman Series
The Lyman Series is an exciting group of spectral lines in the hydrogen spectrum. These lines result from electron transitions ending in the first energy level (n=1). This means electrons are falling from a higher level down to n=1. These transitions emit ultraviolet light, which is not visible to the naked eye.
For example:
  • The first line results from an electron dropping from n=2 to n=1.
  • The second line is from n=3 to n=1.
  • The third line happens when falling from n=4 to n=1.
The Rydberg formula helps calculate the wavelengths of these transitions. For the Lyman series, the formula sets n鈧 = 1 and varies n鈧 for different lines. The result is the characteristic wavelengths of this series.
Balmer Series
The Balmer Series is famous for its bright spectral lines. These occur when electrons transition to the second energy level (n=2). Unlike the Lyman Series, the Balmer Series includes visible light, which falls within the human eye's ability to see, making it special for historical spectroscopy.
Important transitions include:
  • From n=3 to n=2, giving the first line in the series.
  • From n=4 to n=2, resulting in the second line.
  • From n=5 to n=2 for the third line.
Using the Rydberg formula, you can find the exact wavelengths by setting n鈧 = 2 and adjusting n鈧 accordingly. This calculation provides the key to understanding the beautiful colors in this series.
Paschen Series
In the realm of the infrared spectrum, the Paschen Series stands out. It involves transitions where electrons land on the third energy level (n=3), emitting wavelengths primarily in the infrared region.
Key transitions are:
  • The first line from n=4 to n=3.
  • The second line when falling from n=5 to n=3.
  • The third line due to a drop from n=6 to n=3.
To calculate these wavelengths using the Rydberg formula, you choose n鈧 = 3 and vary n鈧 for the different lines. This computation shows why the Paschen Series is not observable without special equipment, as it is outside the visible light range.
Energy Transitions
Energy Transitions are at the heart of spectral line series, linking them to electron movement between energy levels in atoms. When an electron changes its level, either gaining or losing energy, light is absorbed or emitted, forming spectral lines. These changes are discrete, meaning electrons jump from one energy level to another without stopping in-between.
The difference in energy between these two levels dictates the photon energy and thus the wavelength observed. This principle explains why specific wavelengths correspond to different series like Lyman, Balmer, and Paschen.
The Rydberg formula captures this concept by correlating the initial (n鈧) and final (n鈧) quantum numbers to calculate the spectral lines tied to these transitions.
Wavelength Calculation
Wavelength Calculation using the Rydberg formula is straightforward yet fascinating. It requires the known constant, the Rydberg constant (R鈧), and a pair of quantum numbers: n鈧 and n鈧.
The formula:\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\]To find the wavelength (位), solve this equation for different series by plugging in specific values of n鈧 and n鈧. Remember:
  • Lyman Series sets n鈧 to 1.
  • Balmer Series uses n鈧 of 2.
  • Paschen Series implements n鈧 as 3.
As n鈧 increases from n鈧 + 1 upwards, each calculated wavelength corresponds to a spectral line in the series. The series limit is reached as n鈧 approaches infinity, resulting in the shortest possible wavelength.

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Most popular questions from this chapter

The Lyman series in the hydrogen atom corresponds to transitions that originate from the \(n=1\) level in absorption or that terminate in the \(n=1\) level for emission. Calculate the energy, frequency (in inverse seconds and inverse centimeters), and wavelength of the least and most energetic transition in this series.

List the allowed quantum numbers \(m_{l}\) and \(m_{s}\) for the following subshells and determine the maximum occupancy of the subshells: a. \(2 p\) b. \(3 d\) c. \(4 f\) d. \(5 g\)

The inelastic mean free path of electrons in a solid, \(\lambda,\) governs the surface sensitivity of techniques such as AES and XPS. The electrons generated below the surface must make their way to the surface without losing energy in order to give elemental and chemical shift information. An empirical expression for elements that give \(\lambda\) as a function of the kinetic energy of the electron generated in AES or XPS is \(\lambda=538 E^{-2}+0.41(l E)^{0.5} .\) The units of \(\lambda\) are monolayers, \(E\) is the kinetic energy of the electron in eV, and \(l\) is the monolayer thickness in nanometers. On the basis of this equation, what kinetic energy maximizes the surface sensitivity for a monolayer thickness of \(0.3 \mathrm{nm} ?\) An equation solver would be helpful in obtaining the answer.

The absorption spectrum of the hydrogen atom shows lines at \(5334,7804,9145,9953,\) and \(10,478 \mathrm{cm}^{-1} .\) There are no lower frequency lines in the spectrum. Use the graphical methods discussed in Example Problem 22.6 to determine \(n_{\text {initial }}\) and the ionization energy of the hydrogen atom in this state. Assume values for \(n_{\text {initial}}\) of \(1,2,\) and 3.

Two angular momenta with quantum numbers \(j_{1}=3 / 2\) and \(j_{2}=5 / 2\) are added. What are the possible values of \(J\) for the resultant angular momentum states?

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