Problem 44 - Chemistry: The Central Science

Question

The hydrogen atom can absorb light of wavelength \(1094 \mathrm{~nm}\). (a) In what region of the electromagnetic spectrum is this absorption found? (b) Determine the initial and final values of \(n\) associated with this absorption.

Step-by-Step Solution

Answer

The absorption of light with a wavelength of 1094 nm occurs in the infrared region of the electromagnetic spectrum. Using the Rydberg formula and testing integer values for the initial and final principal quantum numbers \(n_1\) and \(n_2\), we find that the initial value is \(n_1 = 1\) and the final value is \(n_2 = 3\).

Step 1: Identify the region in the electromagnetic spectrum

Look up the wavelength intervals for each region of the spectrum to see where 1094 nm falls. The electromagnetic spectrum can be divided into these regions: 1. Radio waves: greater than 1 mm 2. Microwaves: 1 mm - 100 µm 3. Infrared: 100 µm - 750 nm 4. Visible light: 750 nm - 400 nm 5. Ultraviolet: 400 nm - 10 nm 6. X-rays: 10 nm - 0.1 nm 7. Gamma rays: less than 0.1 nm 1094 nm falls within the range of infrared light, so this absorption occurs in the infrared region of the spectrum.

Step 2: Determine the initial and final values of n using the Rydberg formula

The Rydberg formula is given by: \[\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\] where \(\lambda\) is the wavelength, \(R_H\) is the Rydberg constant for hydrogen (\(1.097373 \times 10^7 \mathrm{m^{-1}}\)), and \(n_1\) and \(n_2\) are the initial and final principal quantum numbers, respectively. Since the hydrogen atom absorbs light, we know that an electron transitions to a higher energy level, so \(n_2 > n_1\). First, let's convert the wavelength to meters and calculate the reciprocal: \[\lambda = 1094 \mathrm{~nm} = 1.094 \times 10^{-6} \mathrm{m}\] \[\frac{1}{\lambda} = \frac{1}{1.094 \times 10^{-6}} = 9.137 \times 10^6 \mathrm{m^{-1}}\] Now, plug in the known values into the Rydberg formula: \[(9.137 \times 10^6) = (1.097373 \times 10^7) \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\] Divide by the Rydberg constant so that we can solve for the difference in \(n_1^2\) and \(n_2^2\): \[\frac{9.137 \times 10^6}{1.097373 \times 10^7} = \frac{1}{n_1^2} - \frac{1}{n_2^2}\] \[0.833 = \frac{1}{n_1^2} - \frac{1}{n_2^2}\] Now we need to test different integer values of \(n_1\) and \(n_2\) to find a pair of numbers that satisfies this equation. After testing, we find that: \[0.833 \approx \frac{1}{1^2} - \frac{1}{3^2} = \frac{8}{9}\] So, the initial value, \(n_1\), is 1, and the final value, \(n_2\), is 3. This electron transition corresponds to the absorption of light with a wavelength of 1094 nm.

Key Concepts

Electromagnetic Spectrum

The electromagnetic spectrum is a range that includes all types of electromagnetic radiation. Each type of radiation differs based on wavelength and energy.
The range is vast, covering energy forms from very low-energy radio waves to high-energy gamma rays. Here’s a quick breakdown of the spectrum:

• Radio waves: These have the longest wavelengths, typically greater than 1 mm.
• Microwaves: Shorter than radio waves, ranging from 1 mm to 100 µm.
• Infrared: These have wavelengths between 100 µm and 750 nm, which make them invisible to the naked eye.
• Visible light: The only portion visible to humans, with wavelengths between 750 nm and 400 nm.
• Ultraviolet: Invisible rays between 400 nm and 10 nm, capable of causing sunburn.
• X-rays and Gamma rays: These have very short wavelengths, less than 10 nm, and carry high energy, useful in medical imaging.

Understanding where electromagnetic radiation like light fits into this spectrum helps us determine its properties and potential applications.

Rydberg Formula

The Rydberg formula is a critical tool in understanding the atomic spectra, especially for hydrogen. It helps to calculate the wavelength of light emitted or absorbed when an electron transitions between different energy levels of an atom.
The formula is expressed as: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

• \(\lambda\) is the wavelength of the absorbed or emitted light.
• \(R_H\) is the Rydberg constant for hydrogen, \(1.097373 \times 10^7\, \mathrm{m}^{-1}\).
• \(n_1\) and \(n_2\) are the principal quantum numbers, representing the initial and final energy levels of the electron.

For instance, when a hydrogen electron absorbs a photon, it jumps from a lower to a higher energy level. This concept is vital in understanding why hydrogen emits or absorbs light at specific wavelengths. By using observed wavelengths, we can predict electron transitions and gain insights into atomic structure.

Infrared Region

The infrared region of the electromagnetic spectrum ranges from wavelengths of 750 nm to 1 mm. It is a part of the spectrum that is invisible to the naked eye but can be felt as heat.
This region extends itself more into the light spectrum than visible light does and consists of several subranges based on wavelength:

• Near Infrared: Closest to visible light, with wavelengths from 750 nm to 1400 nm.
• Mid Infrared: Wavelengths range from 1400 nm to 3000 nm, where many thermal sensors operate.
• Far Infrared: Goes from 3000 nm to 1 mm, mostly associated with thermal radiation.

In our specific example, the hydrogen atom absorbs light with a wavelength of 1094 nm, placing it in the near-infrared subrange. Infrared radiation finds extensive use in various technologies such as night-vision equipment, remote control devices, and fiber optic communications. Understanding this region enhances our grasp of how atoms and molecules interact with heat and light.