Problem 36 - Chemistry The Molecular Science

Question

The Brackett series of emissions has \(n_{\mathrm{f}}=4\). (a) Calculate the wavelength, in nanometers, of the photon emitted by the \(n=7\) to \(n=4\) transition. (b) In what region of the electromagnetic spectrum is the emitted radiation?

Step-by-Step Solution

Answer

(a) \(2167 \, \text{nm}\); (b) Infrared region.

Step 1: Understand the Brackett Series

The Brackett series in hydrogen corresponds to electron transitions where the final quantum level is \(n_f = 4\). Photons are emitted when an electron transitions from a higher energy level to a lower one, in this case, from \(n = 7\) to \(n = 4\).

Step 2: Use the Rydberg Formula for Wavelength Calculation

The Rydberg formula for calculating the wavelength \( \lambda \) in the Brackett series is given by: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n^2} \right) \] where \( R_H \) is the Rydberg constant \(1.097 \times 10^7 \, \text{m}^{-1}\). Here, \(n_f = 4\) and \(n = 7\).

Step 3: Substitute Values and Solve for Wavelength

Substitute \( n_f = 4 \) and \( n = 7 \) into the Rydberg formula: \[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4^2} - \frac{1}{7^2} \right) \] Simplify the expression: \[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{16} - \frac{1}{49} \right) \] \[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( 0.0625 - 0.0204 \right) \] \[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.0421 \] \[ \frac{1}{\lambda} = 461317 \] Finally, \( \lambda = \frac{1}{461317} \, \text{m} = 2.167 \times 10^{-6} \, \text{m} \). Convert to nanometers: \( \lambda = 2167 \, \text{nm} \).

Step 4: Determine the Electromagnetic Spectrum Region

The wavelength calculated is \(2167 \, \text{nm}\). The infrared region of the electromagnetic spectrum includes wavelengths from approximately \(700 \, \text{nm}\) to \(1 \, \text{mm}\). Therefore, \(2167 \, \text{nm}\) places the photon in the infrared region.

Key Concepts

Rydberg Formula

The Rydberg Formula is essential for understanding the emission or absorption of light in hydrogen-like atoms. It helps calculate the wavelengths of photons emitted during electron transitions between energy levels. The formula is written as: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n^2} \right) \]Where:

• \( \lambda \) is the wavelength of the emitted light.
• \( R_H \) is the Rydberg constant for hydrogen, typically valued at \(1.097 \times 10^7 \text{ m}^{-1}\).
• \( n_f \) is the final energy level, and \( n \) is the initial energy level.

To find the wavelength of light in a hydrogen transition, insert the respective energy levels into the equation. This formula clarifies how the energy difference between levels dictates the wavelength of light emitted, giving insight into the atomic structure. It's a critical aspect of quantum mechanics, showcasing how electron transitions lead to distinct spectral lines.

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, ranging from very short gamma rays to very long radio waves. Each type of radiation is categorized by its wavelength, frequency, or energy.

• Gamma Rays: Shortest wavelength, high energy.
• X-rays: Longer than gamma rays, but still quite short.
• Ultraviolet: Invisible to the naked eye, longer than X-rays.
• Visible Light: The only part human eyes can see, from violet to red.
• Infrared: Longer than visible light, ranging from \(700 \text{ nm}\) to \(1 \text{ mm}\).
• Microwaves: Longer still, can extend to a few centimeters.
• Radio Waves: The longest wavelength, used for communication.

Understanding where a particular wavelength falls within this spectrum helps us determine the type of electromagnetic radiation. For instance, the \(2167\text{ nm}\) wavelength from our example lands it in the infrared region, indicating it has properties typical of that range.

Hydrogen Emission Spectrum

The hydrogen emission spectrum is a series of lines seen when the light emitted by hydrogen atoms is separated into its component wavelengths. Each line corresponds to a transition between energy levels, as electrons drop from higher to lower levels.In hydrogen, several named series categorize these transitions:

• Lyman Series: Transitions from higher levels to \( n_f = 1 \), in the ultraviolet range.
• Balmer Series: Moves to \( n_f = 2 \), visible and near ultraviolet.
• Paschen Series: Falls to \( n_f = 3 \), infrared region.
• Brackett Series: Falls to \( n_f = 4 \), deeper into infrared.
• Pfund and Humphreys Series: Descend further to \( n_f = 5 \) and \( n_f = 6 \), respectively.

These spectral lines provide invaluable insights into the atomic structure and quantum mechanics. The Brackett series specifically, seen in the infrared, reflects less energetic transitions compared to those in visible or ultraviolet ranges. Through studying these spectral lines, scientists could ascertain information about energy levels, transition probabilities, and even deduce aspects of star compositions.