Problem 136 - Chemistry An Atoms First Approach

Question

A carbon-oxygen double bond in a certain organic molecule absorbs radiation that has a frequency of \(6.0 \times 10^{13} \mathrm{s}^{-1}\). a. What is the wavelength of this radiation? b. To what region of the spectrum does this radiation belong? c. What is the energy of this radiation per photon? d. A carbon-oxygen bond in a different molecule absorbs radiation with frequency equal to \(5.4 \times 10^{13} \mathrm{s}^{-1} .\) Is this radiation more or less energetic?

Step-by-Step Solution

Answer

a. The wavelength of the radiation is \(5.00\ \mu m\). b. The radiation belongs to the infrared region of the spectrum. c. The energy of the radiation per photon is \(3.98 \times 10^{-20}\ J\). d. The radiation with frequency \(6.0 \times 10^{13}\ s^{-1}\) is more energetic compared to the one with frequency \(5.4 \times 10^{13}\ s^{-1}\).

Step 1: Calculate the wavelength of the radiation

Given the frequency \(\nu = 6.0 \times 10^{13}\ s^{-1}\), we can find the wavelength \(\lambda\) using the equation \(c = \lambda \nu\), where \(c\) is the speed of light, which is approximately \(3.00 \times 10^8\ m/s\). To find \(\lambda\), we can rearrange the equation to get \(\lambda = \frac{c}{\nu}\), and then plug in the given values: \(\lambda = \frac{3.00 \times 10^8\ m/s}{6.0 \times 10^{13}\ s^{-1}} = 5.00 \times 10^{-6}\ m = 5.00\ \mu m\) The wavelength of the radiation is \(5.00\ \mu m\).

Step 2: Determine the region of the spectrum

The wavelength \(5.00\ \mu m\) falls within the infrared (IR) region of the electromagnetic spectrum. So, the radiation belongs to the infrared region.

Step 3: Calculate the energy per photon

To find the energy per photon, we can use the energy equation \(E = h\nu\), where \(h\) is the Planck's constant (\(6.626 \times 10^{-34}\ Js\)) and \(\nu\) is the frequency. \(E = (6.626 \times 10^{-34}\ Js)(6.0 \times 10^{13}\ s^{-1}) = 3.98 \times 10^{-20}\ J\) The energy of the radiation per photon is \(3.98 \times 10^{-20}\ J\).

Step 4: Compare the energies of the two radiations

To compare the energies of the two radiations, we can calculate the energy of the radiation with the frequency of \(5.4 \times 10^{13}\ s^{-1}\) using the same energy equation: \(E = (6.626 \times 10^{-34}\ Js)(5.4 \times 10^{13}\ s^{-1}) = 3.58 \times 10^{-20}\ J\) Since the energy of the radiation with frequency \(6.0 \times 10^{13}\ s^{-1}\) (\(3.98 \times 10^{-20}\ J\)) is larger than the energy of the radiation with frequency \(5.4 \times 10^{13}\ s^{-1}\) (\(3.58 \times 10^{-20}\ J\)), the radiation with frequency \(6.0 \times 10^{13}\ s^{-1}\) is more energetic.

Key Concepts

Wavelength Calculation

Wavelength is a crucial parameter when discussing electromagnetic radiation. It's represented by the Greek letter \( \lambda \). Wavelength measures the distance between two consecutive peaks or troughs in a wave. To calculate wavelength when you have the frequency, you can use the equation \( c = \lambda u \). Here, \( c \) represents the speed of light which is approximately \( 3.00 \times 10^8 \ m/s \), \( \lambda \) is the wavelength, and \( u \) is the frequency of the wave. By rearranging this equation, you can find wavelength as \( \lambda = \frac{c}{u} \).
For instance, if the frequency of radiation absorbed by a carbon-oxygen double bond is \( 6.0 \times 10^{13} \ s^{-1} \), you would substitute it into the equation along with the speed of light to find \( \lambda \) as \( 5.00 \times 10^{-6} \ m \) or \( 5.00 \ \mu m \). This calculation is essential for understanding where this radiation lands on the electromagnetic spectrum.
Calculating wavelength is not just about solving an equation. It helps you visualize the wave and understand its interaction with matter. Shorter wavelengths tend to carry more energy, which becomes crucial in applications like infrared technologies and other electromagnetic spectrum uses.

Infrared Radiation

Infrared (IR) radiation is a type of electromagnetic radiation with wavelengths longer than visible light but shorter than microwaves, typically in the range of \( 1 \mu m \) to \( 1000 \mu m \). This type of radiation is generally associated with heat.
When a carbon-oxygen double bond absorbs IR radiation, it typically involves wavelengths like \( 5.00 \mu m \), as calculated for a specific case, placing it firmly within the infrared zone of the electromagnetic spectrum. The absorption of IR radiation causes molecules to vibrate, which is why infrared radiation is often studied in the context of molecular bonds and structures. IR spectroscopy uses this property to identify and analyze molecular compositions in various scientific fields.
Understanding IR is essential not only in fields such as chemistry and astronomy but also in everyday applications like remote controls and even human body temperature detection. Its role in absorbing, emitting, and transmitting heat makes IR a fundamental concept in thermal imaging, allowing for diverse applications across multiple industries.

Photon Energy

Photon energy is a concept that explains the energy carried by a single photon of electromagnetic radiation, commonly calculated using Planck's equation \( E = hu \). In this equation, \( E \) represents the energy of the photon, \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \ Js \)), and \( u \) is the frequency of the radiation.
In the specific example of a carbon-oxygen double bond absorbing radiation with a frequency of \( 6.0 \times 10^{13} \ s^{-1} \), the photon energy is calculated as \( 3.98 \times 10^{-20} \ J \). This determines how much energy is involved when a molecule absorbs a photon, which is a critical aspect of understanding molecular interactions and reactions.
Moreover, the comparison of photon energies between different frequencies shows that a higher frequency (like \( 6.0 \times 10^{13} \ s^{-1} \)) leads to higher energy photons compared to a lower frequency, emphasizing the direct relationship between frequency and energy. Understanding this relationship is vital for fields such as quantum mechanics and spectroscopy, where the behavior and energy levels of small particles are analyzed. The mastery of photon energy concepts enlightens a vast array of topics, including solar energy technology and the fundamental principles of light.