Question

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

(a) The region of the electromagnetic spectrum for the Lyman series is defined by the minimum and maximum wavelengths. The minimum wavelength (\(\lambda_{min}\)) is \(\frac{1}{1.097 \times 10^7 m^{-1}} \approx 9.11 \times 10^{-8} m\), and the maximum wavelength (\(\lambda_{max}\)) is \(\frac{4}{3(1.097 \times 10^7 m^{-1})} \approx 1.22 \times 10^{-7} m\). This range corresponds to the ultraviolet region of the electromagnetic spectrum. (b) The first three wavelengths in the Lyman series are: - For \(n_i = 2\): \( \lambda_1 = \frac{4}{3 R_H} \approx 1.22 \times 10^{-7} m\) - For \(n_i = 3\): \( \lambda_2 = \frac{9}{8 R_H} \approx 1.03 \times 10^{-7} m\) - For \(n_i = 4\): \( \lambda_3 = \frac{16}{15 R_H} \approx 9.76 \times 10^{-8} m\)

Step-by-step solution

Determine the region of the electromagnetic spectrum for Lyman series

According to the question, for the Lyman series, \(n_f\) is always equal to 1. The electromagnetic spectrum region could be determined by finding the minimum and maximum possible wavelength values. For the minimum wavelength, let's consider the transition for which \(n_i\) approaches infinity: \( \lim_{n_i \to \infty} \frac{1}{\lambda_{min}} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \) \( \lim_{n_i \to \infty} \frac{1}{\lambda_{min}} = R_H \left( 1 - 0 \right) = R_H \) So, \( \lambda_{min} = \frac{1}{R_H} = \frac{1}{1.097 \times 10^7 m^{-1}} \) For the maximum wavelength, let's consider the transition for which \(n_i = 2 \): \( \frac{1}{\lambda_{max}} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R_H \) So, \( \lambda_{max} = \frac{4}{3 R_H} = \frac{4}{3(1.097 \times 10^7 m^{-1})} \) Now, calculate the values of \( \lambda_{min} \) and \( \lambda_{max} \).

Calculate the wavelengths of the first three lines in the Lyman series

To find the wavelengths for the first three lines of the Lyman series, we will plug the values of \(n_i\) (2, 3, and 4) into the Rydberg formula. For \(n_i = 2\): \( \frac{1}{\lambda_1} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R_H \) For \(n_i = 3\): \( \frac{1}{\lambda_2} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) = R_H \left( 1 - \frac{1}{9} \right) = \frac{8}{9} R_H \) For \(n_i = 4\): \( \frac{1}{\lambda_3} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) = R_H \left( 1 - \frac{1}{16} \right) = \frac{15}{16} R_H \) Finally, calculate the values of \( \lambda_1 \), \( \lambda_2 \), and \( \lambda_3 \).

Key concepts

Hydrogen Atom

The hydrogen atom is the simplest and most fundamental atom, consisting of just one proton and one electron. Despite its simplicity, it's a vital building block in understanding atomic structure. When energy is added to a hydrogen atom, its electron can jump to a higher energy level or orbit. This excited state, however, is unstable, and the electron will eventually fall back to a lower energy level, emitting a photon—a particle of light—as it does so.
This emitted light forms distinct spectral lines, which are characteristic of the electron transitions within the atom. These lines are grouped into series depending on the final energy level of the electron. The Lyman series, for instance, is one such group where the electron falls to the ground state (n=1).

Rydberg Formula

The Rydberg formula is crucial in calculating the wavelengths of the spectral lines of hydrogen. It mathematically describes the relationship between the wavelengths of emitted or absorbed light and the energy levels of an electron within an atom. The formula is:\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]where \(\lambda\) is the wavelength, \(R_H\) is the Rydberg constant approximately equal to \(1.097 \times 10^7 \text{m}^{-1}\), \(n_f\) is the final energy level, and \(n_i\) is the initial energy level.
For the Lyman series, \(n_f\) is always 1, meaning the electron is moving to the first energy level (ground state). By substituting different values of \(n_i\) (e.g., 2, 3, or 4) into this equation, students can calculate the different wavelengths of light emitted during these transitions.

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, from radio waves to gamma rays. Each type of radiation is characterized by its wavelength and frequency. The visible spectrum, which our eyes can detect, represents only a small portion of this spectrum.
The Lyman series of the hydrogen atom involves transitions that produce ultraviolet light, which lies in the part of the electromagnetic spectrum with shorter wavelengths than visible light. When looking at the wavelengths of the Lyman series, they fall within the ultraviolet region, which is not visible to the naked eye but is important in various scientific applications.

Wavelength Calculation

Calculating the exact wavelengths for transitions in the hydrogen atom, such as those in the Lyman series, requires careful application of the Rydberg formula. To do this:

• Identify the initial (\(n_i\)) and final (\(n_f\)) energy levels for the electron.
• Use \(n_f = 1\) for all Lyman series calculations.
• Substitute the values of \(n_i\) into the formula.

For example, the first three lines of the Lyman series occur when the electron transitions from \(n_i = 2, 3,\) and \(4\) to \(n_f = 1\). Calculating these gives:

• For \(n_i = 2\), \( \lambda_1 = \frac{4}{3 R_H} \)
• For \(n_i = 3\), \( \lambda_2 = \frac{9}{8 R_H} \)
• For \(n_i = 4\), \( \lambda_3 = \frac{16}{15 R_H} \)

By substituting \(R_H\), these expressions can determine precise wavelengths, showcasing the energy changes within the hydrogen atom.