Question

The Rydberg - Ritz equation governing the spectral lines of hydrogen is \((1 / \lambda)=\mathrm{R}\left[\left(1 / \mathrm{n}_{1}{ }^{2}\right)-\left(1 / \mathrm{n}_{2}{ }^{2}\right)\right]\), where \(\mathrm{R}\) is the Rydberg constant, \(\mathrm{n}_{1}\) indexes the series under consideration \(\left(\mathrm{n}_{1}=1\right.\) for the Lyman series, \(\mathrm{n}_{1}=2\) for the Balmer series, \(\mathrm{n}_{1}=3\) for the Paschen series \(), \mathrm{n}_{2}=\mathrm{n}_{1}+1, \mathrm{n}_{1}+2, \mathrm{n}_{1}+3, \ldots\) indexes the successive lines in a series, and \(\lambda\) is the wave- length of the line corresponding to index \(\mathrm{n}_{2}\). Thus, for the Lyman series, \(\mathrm{n}_{1}=1\) and the first two lines are \(1215.56 \AA\left(\mathrm{n}_{2}=\mathrm{n}_{1}+1=2\right)\) and \(1025.83 \AA\left(\mathrm{n}_{2}=\mathrm{n}_{1}+2=3\right)\) Using these two lines, calculate two separate values of the Rydberg constant. The actual value of this constant is \(\mathrm{R}=109678 \mathrm{~cm}^{-1}\)

Short answer

The Rydberg constant is calculated for the first line (λ₁ = 1215.56 Å, n₂ = 2) and the second line (λ₂ = 1025.83 Å, n₂ = 3) using the Rydberg-Ritz equation. The results are as follows: R₁ = 109696.058 cm⁻¹ and R₂ = 109747.995 cm⁻¹. These values are close to the actual Rydberg constant value, R = 109678 cm⁻¹, demonstrating the usefulness of the Rydberg-Ritz equation for obtaining spectral lines of hydrogen.

Step-by-step solution

Rearrange the equation for the Rydberg constant

First, we need to solve the Rydberg-Ritz equation for the Rydberg constant (R). The equation is given as: \[\frac{1}{\lambda} = R\left[\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right]\] Rearranging the equation for R: \[R = \frac{1}{\lambda} \cdot \frac{1}{\left[\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right]}\]

Calculate the Rydberg constant for the first line

Now we will substitute the given values for the first line (λ₁ = 1215.56 Å and n₂ = 1 + 1 = 2) to find the first value of the Rydberg constant R₁: \[R_1 = \frac{1}{1215.56\cdot 10^{-8}} \cdot \frac{1}{\left[\frac{1}{1^{2}} - \frac{1}{2^{2}}\right]} = \frac{1}{1.21556\cdot 10^{-5}} \cdot \frac{1}{\left[1 - \frac{1}{4}\right]}\] Calculate the result: \[R_1 = \frac{1}{1.21556\cdot 10^{-5}} \cdot \frac{1}{\frac{3}{4}} = 109696.058\,\text{cm}^{-1}\]

Calculate the Rydberg constant for the second line

Next, we will substitute the given values for the second line (λ₂ = 1025.83 Å and n₂ = 1 + 2 = 3) to find the second value of the Rydberg constant R₂: \[R_2 = \frac{1}{1025.83\cdot 10^{-8}} \cdot \frac{1}{\left[\frac{1}{1^{2}} - \frac{1}{3^{2}}\right]} = \frac{1}{1.02583\cdot 10^{-5}} \cdot \frac{1}{\left[1 - \frac{1}{9}\right]}\] Calculate the result: \[R_2 = \frac{1}{1.02583\cdot 10^{-5}} \cdot \frac{1}{\frac{8}{9}} = 109747.995\,\text{cm}^{-1}\]

Compare the calculated values with the actual value

Now, let's compare the calculated values of the Rydberg constant for the first and second lines with the actual value provided: Actual Value: R = 109678 cm⁻¹ Calculated Value (First Line): R₁ = 109696.058 cm⁻¹ Calculated Value (Second Line): R₂ = 109747.995 cm⁻¹ The calculated values are close to the actual value of the Rydberg constant, demonstrating the use of the Rydberg-Ritz equation to determine spectral lines of hydrogen.

Key concepts

Rydberg Constant

The Rydberg constant (\( R \)) is a fundamental physical constant pivotal in the study of atomic physics and spectroscopy. It is named after the Swedish physicist Johannes Rydberg, who devised the Rydberg formula to describe the wavelengths of spectral lines of many chemical elements.

The constant is specifically vital in the analysis of the spectral lines of hydrogen, the simplest atom. It represents the limit of the highest wavenumber (inverse wavelength) of any photon that can be emitted from the hydrogen atom, or equivalently, from a hydrogen-like ion.

When dealing with the Rydberg constant, it is crucial to consider units for accuracy; it is often expressed in per centimeter (\( \text{cm}^{-1} \)). The recognized value of the Rydberg constant is approximately \( 109,678 \text{cm}^{-1} \), which serves as a benchmark for calculating and comparing observed spectral lines.

Spectral Lines of Hydrogen

The spectral lines of hydrogen are a manifestation of electronic transitions between different energy levels in a hydrogen atom. According to the Bohr model, when an electron transits from a higher energy level (\( n_2 \)) to a lower energy level (\( n_1 \)), it emits radiation with specific wavelength, creating what we see as a spectral line.

These lines are grouped into series named after the scientists who discovered them: the Lyman, Balmer, Paschen, Brackett, and Pfund series, each corresponding to transitions ending at different base energy levels (\( n_1 \)). For example, the Lyman series corresponds to transitions where the final energy level is 1, and the Balmer to where it's 2.

Understanding these series is critical to learning about hydrogen's atomic structure and the electronic configurations of its atom. The positions of these lines on the electromagnetic spectrum offer great insight into the energy levels within the atom and are instrumental in various applications such as astrophysics, quantum mechanics, and chemistry.

Wavelength Calculation

Wavelength calculation is a fundamental aspect of optics and quantum physics, involving determining the distance between repeating units of a wave pattern. It can be exemplified by calculating the wavelengths of light emitted during electron transitions as seen in spectral lines.

Using the Rydberg-Ritz equation for the spectral lines of hydrogen, we can calculate the wavelength (\( \lambda \)) of emitted photons during electronic transitions. The equation is as follows:\[(1 / \lambda)= R[ (1 / n_1^{2}) - (1 / n_2^{2}) ]\]
Where \( R \) is the Rydberg constant, \( n_1 \) and \( n_2 \) are integers representing the principal quantum numbers of the electrons' initial and final orbits, respectively.

In wavelength calculations, the given values or measured spectral lines can be substituted into the equation to solve for the observed wavelengths or the corresponding Rydberg constant. It is also important to note that the values of \( \lambda \) obtained help us to understand the electronic structure of the atom and its energy transitions, which have a broad range of practical applications, such as in the development of lasers, fluorescent lights, and understanding astronomical phenomena.