Problem 42 - Chemistry: The Central Science in SI Units, Global Edition

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 \(n_{1}=2,3,\) and \(4 .\)

Step-by-Step Solution

Answer

(a) The Lyman series is observed in the ultraviolet region of the electromagnetic spectrum. (b) The first three wavelengths in the Lyman series are approximately \(1.215 \times 10^{-7}\,m\), \(1.025 \times10^{-7}\,m\), and \(9.733 \times 10^{-8}\,m\).

Step 1: Understand and write the Rydberg formula for hydrogen emission lines.

The Rydberg formula for the hydrogen atom emission lines is given by: \[\frac{1}{\lambda}=R_H\left(\frac{1}{n_{f}^2}-\frac{1}{n_{i}^2}\right)\] where: - \(\lambda\) is the wavelength of the emitted light; - \(R_H\) is the Rydberg constant for hydrogen, approximately equal to \(1.097 \times 10^7\,m^{-1}\); - \(n_f\) is the final energy level of the electron; - \(n_i\) is the initial energy level of the electron. For the Lyman series, \(n_f = 1\).

Step 2: Calculate the wavelengths of the first three lines in the Lyman series.

Now, we will calculate the wavelengths of the first three lines in the Lyman series, i.e., for \(n_i = 2, 3\), and \(4\). Using the Rydberg formula for each value, we get: For \(n_i = 2\): \[\frac{1}{\lambda_1}=R_H\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\] \[\lambda_1=\frac{1}{R_H\left(\frac{3}{4}\right)}\] For \(n_i = 3\): \[\frac{1}{\lambda_2}=R_H\left(\frac{1}{1^2}-\frac{1}{3^2}\right)\] \[\lambda_2=\frac{1}{R_H\left(\frac{8}{9}\right)}\] For \(n_i = 4\): \[\frac{1}{\lambda_3}=R_H\left(\frac{1}{1^2}-\frac{1}{4^2}\right)\] \[\lambda_3=\frac{1}{R_H\left(\frac{15}{16}\right)}\] We can now calculate the wavelengths numerically for each case: \[\lambda_1 \approx \frac{1}{1.097\times10^{7}\left(\frac{3}{4}\right)} = 1.215\times10^{-7}\,m\] \[\lambda_2 \approx \frac{1}{1.097\times10^{7}\left(\frac{8}{9}\right)} = 1.025\times10^{-7}\,m\] \[\lambda_3 \approx \frac{1}{1.097\times10^{7}\left(\frac{15}{16}\right)} = 9.733\times10^{-8}\,m\]

Step 3: Determine the region of the electromagnetic spectrum in which the Lyman series is observed.

Now that we have the wavelengths of the first three lines of the Lyman series, we can determine the region of the electromagnetic spectrum in which these lines are observed. The ranges for different types of electromagnetic radiation are as follows: - Radio waves: \(\lambda > 10^{-1} \,m\) - Microwaves: \(10^{-1}\,m > \lambda > 10^{-3}\,m\) - Infrared: \(10^{-3}\,m > \lambda > 7 \times 10^{-7}\,m\) - Visible light: \(7 \times 10^{-7}\,m > \lambda > 4 \times 10^{-7}\,m\) - Ultraviolet: \(4 \times 10^{-7}\,m > \lambda > 10^{-8}\,m\) - X-rays: \(10^{-8}\,m > \lambda > 10^{-11}\,m\) - Gamma rays: \(\lambda < 10^{-11}\,m\) As we can see from the calculated values, the first three lines of the Lyman series fall within the ultraviolet range (specifically, all three wavelengths are between \(4 \times 10^{-7}\,m\) and \(10^{-8}\,m\). So the final answers are: (a) The Lyman series is observed in the ultraviolet region of the electromagnetic spectrum. (b) The first three wavelengths in the Lyman series are approximately \(1.215 \times 10^{-7}\,m\), \(1.025 \times10^{-7}\,m\), and \(9.733 \times 10^{-8}\,m\).

Key Concepts

Rydberg Formula

The Rydberg formula is a crucial mathematical expression used to predict the wavelengths of light emitted or absorbed by electrons transitioning between energy levels within an atom, particularly hydrogen. It serves as a cornerstone for understanding the spectral lines of the hydrogen atom. The formula is expressed as:

• \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right) \)

Here's what each component represents:

• \(\lambda\) is the wavelength of the emitted light.
• \(R_H\) stands for the Rydberg constant, which is approximately \(1.097 \times 10^7\,m^{-1}\) for hydrogen.
• \(n_f\) is the final energy level of the electron (for Lyman series, \(n_f = 1\)).
• \(n_i\) is the initial energy level from which the electron transitions.

This formula allows us to calculate the precise wavelengths for different electron transitions, helping in understanding the behavior of atoms when they emit light.

Hydrogen Atom

The hydrogen atom is the simplest atom and consists of only one proton and one electron. Its simplicity makes it a fantastic subject for studying fundamental atomic behavior and quantum mechanics. In terms of spectral lines, hydrogen provides distinct series of lines such as Lyman, Balmer, and Paschen, each corresponding to electrons transitioning to a specific lower energy level.

The Lyman Series

This series involves electron transitions where the final energy level (\(n_f\)) is the first energy level (\(n_1\)). It consists of photons emitted when electrons drop from higher energy levels (such as \(n_2 = 2, 3, 4, \ldots\)) to the first energy state. These transitions generate emissions in the ultraviolet part of the electromagnetic spectrum, which are particularly crucial for studying stellar atmospheres and cosmic rays.

Electromagnetic Spectrum

The electromagnetic spectrum is the range of all types of electromagnetic radiation. Radiation is the means through which energy travels and spreads out. One important aspect of the spectrum is that each type of electromagnetic wave has a different range of wavelengths and frequencies. These include:

• Radio waves: Longest wavelengths, ranging above \(10^{-1}\, m\)
• Microwaves: Between \(10^{-3}\, m\) and \(10^{-1}\, m\)
• Infrared: Range from \(7 \times 10^{-7}\, m\) to \(10^{-3}\, m\)
• Visible light: Spectrally visible colors, between \(4 \times 10^{-7}\, m\) and \(7 \times 10^{-7}\, m\)
• Ultraviolet: Slightly shorter than visible light, \(10^{-8}\, m\) to \(4 \times 10^{-7}\, m\)
• X-rays: Ranging from \(10^{-11}\, m\) to \(10^{-8}\, m\)
• Gamma rays: Shortest wavelengths, less than \(10^{-11}\, m\)

These waves differ in their energy; longer wavelengths have lower energy, while shorter wavelengths are more energetic. The Lyman series falls under the ultraviolet category, which is high-energy and invisible to the naked eye.

Wavelength Calculation

Wavelength calculation is a crucial part of understanding how electrons transition between energy levels. It allows us to infer properties about atomic behavior and the nature of electromagnetic radiation.Using the Rydberg formula, you can calculate the wavelength of light emitted during such transitions. Here's how you would approach the calculation for the Lyman series:

• Insert the known values: For calculating the Lyman series, use \(n_f = 1\) and the appropriate \(n_i\) values (e.g., \(n_i = 2, 3, 4\)).
• Plug the values into the Rydberg formula: \( \frac{1}{\lambda} = 1.097 \times 10^{7}\left(\frac{1}{1^2} - \frac{1}{n_{i}^2}\right) \)
• Solve for \(\lambda\) to get the wavelength.Calculate numerically for each transition to find \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\).

This systematic approach provides the exact wavelengths of the emitted photons, helping to precisely classify the lines within the electromagnetic spectrum and study atomic emissions.