Problem 52 - Chemistry and Chemical Reactivity

Question

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.

Step-by-Step Solution

Answer

The wavelength is approximately 7460 nm and the frequency is approximately \( 4.02 \times 10^{13} \) Hz.

Step 1: Understand the Pfund Series

The Pfund series corresponds to electrons transitioning to the quantum level \( n = 5 \) from higher levels (\( n > 5 \)). The lowest energy transition happens when an electron falls from the next highest level, which is \( n = 6 \), to \( n = 5 \). This transition will give us the longest wavelength within the Pfund series.

Step 2: Use the Rydberg Formula for Wavelength

The wavelength \( \lambda \) of an emitted photon during an electron transition in a hydrogen atom can be calculated using the Rydberg formula:\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where \( R_H = 1.097 \times 10^7 \, \mathrm{m}^{-1} \) is the Rydberg constant, \( n_1 \) is the lower energy level (5 for Pfund), and \( n_2 \) is the higher energy level (6 in this case).

Step 3: Perform Wavelength Calculation

Substitute \( n_1 = 5 \) and \( n_2 = 6 \) into the Rydberg formula:\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{5^2} - \frac{1}{6^2} \right) \]\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{25} - \frac{1}{36} \right) \]Calculate the values of \( \frac{1}{25} = 0.04 \) and \( \frac{1}{36} = 0.0278 \). So,\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times (0.04 - 0.0278) = 1.097 \times 10^7 \times 0.0122 \]\[ \lambda = \frac{1}{1.097 \times 10^7 \times 0.0122} \approx 7.460 \times 10^{-6} \, \mathrm{m} \]Thus, the wavelength is approximately \( 7460 \, \mathrm{nm} \).

Step 4: Calculate the Frequency

Next, use the speed of light \( c = 3.00 \times 10^8 \, \mathrm{m/s} \) to find the frequency \( f \) using the relationship \( c = \lambda f \):\[ f = \frac{c}{\lambda} \]Substitute \( \lambda = 7.460 \times 10^{-6} \, \mathrm{m} \):\[ f = \frac{3.00 \times 10^8}{7.460 \times 10^{-6}} \approx 4.02 \times 10^{13} \, \mathrm{Hz} \].So, the frequency of this line is approximately \( 4.02 \times 10^{13} \, \mathrm{Hz} \).

Key Concepts

Rydberg Formula

The Rydberg formula is a mathematical equation used to predict the wavelengths of the spectral lines of hydrogen atoms. It was introduced by Johannes Rydberg in the late 19th century and has been a cornerstone in the field of atomic physics. The formula calculates the energy difference between two quantum levels during an electron transition in a hydrogen atom, which in turn determines the wavelength of the light emitted. The general form of the Rydberg formula is given as:\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where:

• \( \lambda \): Wavelength of the emitted light.
• \( R_H \): Rydberg constant, approximately \( 1.097 \times 10^7 \, \mathrm{m}^{-1} \).
• \( n_1 \) and \( n_2 \): Principal quantum numbers of the energy levels involved, with \( n_1 < n_2 \).

This formula allows scientists and students to calculate the specific wavelengths associated with different electron transitions, revealing the distinct spectral lines of the hydrogen atom.

Wavelength Calculation

Wavelength calculation is an essential part of understanding how light is emitted or absorbed by atoms. Using the Rydberg formula, we can calculate the wavelength of light emitted when an electron transitions between energy levels in an atom. This is particularly useful for hydrogen, which exhibits a simple spectral line structure.For the Pfund series, the transition of an electron from a higher energy level \( n_2 \) to \( n_1 = 5 \) results in the emission of light in the infrared region. To find the wavelength, we substitute the known values for \( n_1 \) and \( n_2 \) into the Rydberg formula:\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{5^2} - \frac{1}{6^2} \right) \]By calculating the values inside the formula, we derive the wavelength of approximately \( 7460 \, \mathrm{nm} \), placing it in the infrared region of the electromagnetic spectrum.

Quantum Levels

Quantum levels, or energy levels, refer to the fixed energy levels that electrons occupy within an atom. These levels are quantized, meaning electrons can only exist in specific energy states specified by a principal quantum number \( n \). The energy level of an electron is higher when it is further away from the nucleus and lower when it is closer.In hydrogen atoms, when an electron falls from a higher energy state (e.g., \( n = 6 \)) to a lower one (e.g., \( n = 5 \)) in the Pfund series, energy is emitted in the form of light. This concept is fundamental because it explains why we see distinct lines in the atomic spectrum rather than a continuous range of colors. Each transition between specific quantum levels corresponds to the emission of light at a specific wavelength.

Electron Transitions

Electron transitions are the movement of electrons between different energy levels within an atom. These transitions are responsible for the emission or absorption of light in the form of photons. In hydrogen, when an electron jumps from a higher energy level (like \( n = 6 \)) to a lower level (like \( n = 5 \)), it emits a photon whose energy corresponds to the difference in energy between these levels.The energy of the emitted photon can be related to its wavelength through the equation:\[ E = hf = \frac{hc}{\lambda} \]where:

• \( E \): Energy of the photon.
• \( h \): Planck's constant \( 6.626 \times 10^{-34} \, \mathrm{Js} \).
• \( f \): Frequency of the photon.
• \( c \): Speed of light \( 3.00 \times 10^8 \, \mathrm{m/s} \).
• \( \lambda \): Wavelength of the photon.

This relationship helps us understand how different electron transitions determine the frequency and wavelength of the photons emitted, giving us insight into the atomic structure and behavior.