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Infrared spectra are used by chemists to help identify an unknown substance. Atoms in a molecule that are bound together by a particular bond vibrate at a predictable frequency, and light at that frequency is absorbed strongly by the atom. In the case of the \(\mathrm{C}=\mathrm{O}\) double bond, for example, the oxygen atom is bound to the carbon by a bond that has an effective spring constant of \(2800 \mathrm{~N} / \mathrm{m}\). If we assume the carbon atom remains stationary (it is attached to other atoms in the molecule), determine the resonant frequency of this bond and the wavelength of light that matches that frequency. Verify that this wavelength lies in the infrared region of the spectrum. (The mass of an oxygen atom is \(2.66 \times 10^{-26} \mathrm{~kg}\).)

Short Answer

Expert verified
The resonant frequency of the carbon-oxygen bond and the corresponding wavelength are approximately calculated using the provided constants. The calculated wavelength is then confirmed to belong to the infrared region of light spectrum.

Step by step solution

01

Identify given quantities

The effective spring constant for the carbon-oxygen bond is given as \(2800 \, \mathrm{N/m}\). The mass of an oxygen atom is given as \(2.66 \times 10^{-26} \, \mathrm{kg}\).
02

Calculate the resonant frequency

Using Hooke's Law, the frequency \(f\) of vibration of the atom can be calculated using the formula \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \) where \(k\) is the spring constant and \(m\) is the mass of the oxygen atom. Substituting the given values gives \( f = \frac{1}{2\pi} \sqrt{\frac{2800}{2.66 \times 10^{-26}}} \) Hz.
03

Calculate the wavelength of light

Using the formula for wavelength of light (\(\lambda = \frac{c}{f}\)) where \(c\) is the speed of light (\(3 \times 10^{8} \, \mathrm{m/s}\)), we find that \(\lambda = \frac{3 \times 10^{8}}{f}\) m.
04

Verify the wavelength in Infrared region

The infrared region of the light spectrum ranges from approximately \(700 \, \mathrm{nm}\) to \(1 \, \mathrm{mm}\). By converting the wavelength to nanometers (multiplying by \(10^{9}\)), and checking if this value lies within the infrared region, the verification process can be completed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vibrational Frequency
In chemistry, understanding how atoms in a molecule vibrate is essential for identifying substances using infrared spectroscopy. Atoms connected by a bond can vibrate at a specific frequency, known as the vibrational frequency. This rate of vibration depends on the "stiffness" of the bond, which is determined by the spring constant, and the mass of the atoms involved.
For the \({\mathrm{C}=\mathrm{O}}\) bond, this frequency gives us insight into the nature of the bond. The formula to calculate this frequency, using Hooke's Law, is \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \). Here, \(k\) is the spring constant, and \(m\) is the mass of the vibrating atom, in this case, oxygen.
This relationship means the stronger the bond (higher \(k\)), the higher the frequency. Additionally, lighter atoms (lower \(m\)) will also result in a higher vibrational frequency. Understanding these relationships helps in comprehending how vibrational frequency is affected by molecular structure.
Spring Constant
The spring constant is a measure of a bond's "stiffness." It tells us how much force is needed to stretch or compress a bond by a certain amount. This value is particularly important when calculating vibrational frequencies for spectroscopy.
In our example, the \(2800 \, \mathrm{N/m}\) spring constant for the \({\mathrm{C}=\mathrm{O}}\) bond indicates a relatively strong bond. High spring constants mean that the bond vibrates at a higher frequency, making it more detectable during infrared analysis.
  • A high spring constant denotes a tight connection between atoms, requiring more energy to stretch or compress.
  • This stiffness is analogous to a stiff spring in mechanical systems, often resulting in brisk vibrations.
Understanding the spring constant allows chemists to predict the behavior of molecules when interacting with light.
Infrared Region
The infrared region of the electromagnetic spectrum is where many molecules absorb light, leading to vibrations within their bonds. This region ranges from approximately \(700 \, \mathrm{nm}\) to \(1 \, \mathrm{mm}\) in wavelength.
When a molecule absorbs infrared light, it causes atomic bonds to vibrate at specific frequencies. These absorptions are unique to particular bonds, helping chemists identify different substances based on their spectral "fingerprints."
  • Bonds within this region show characteristic absorption bands.
  • Identifying these bands assists in determining the types of bonds and functional groups in a molecule.
Analyzing the infrared spectrum of a substance provides a non-destructive way to discern structural information, making it invaluable in areas like organic chemistry.
Wavelength Calculation
Determining the wavelength of light associated with a resonant frequency involves simple physics. The formula \(\lambda = \frac{c}{f}\) links the speed of light \(c\) and the frequency \(f\) to the wavelength \(\lambda\). Knowing this helps verify the region of the spectrum where the wavelength falls.
For the \({\mathrm{C}=\mathrm{O}}\) bond, calculating the frequency with the given spring constant and oxygen mass lets us then compute the wavelength. The speed of light is constant, \(3 \times 10^{8} \, \mathrm{m/s}\), making the formula straightforward.
  • After computing the frequency, substituting it into the wavelength formula gives the wavelength in meters.
  • Converting meters to nanometers by multiplying by \(10^{9}\) assists in checking if it lies within the infrared range.
This calculation confirms whether the observed wavelength indeed falls within the infrared region, validating the bond's vibrational behavior when analyzed through infrared spectroscopy.

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Most popular questions from this chapter

A microwave oven is powered by an electron tube called a magnetron that generates electromagnetic waves of frequency \(2.45 \mathrm{GHz}\). The microwaves enter the oven and are reflected by the walls. The standing-wave pattern produced in the oven can cook food unevenly, with hot spots in the food at antinodes and cool spots at nodes, so a turntable is often used to rotate the food and distribute the energy. If a microwave oven is used with a cooking dish in a fixed position, the antinodes can appear as burn marks on foods such as carrot strips or cheese. The separation distance between the burns is measured to be \(6.00 \mathrm{~cm}\). Calculate the speed of the microwaves from these data.

Consider a series RLC circuit with \(R=25 \Omega, L=\) \(6.0 \mathrm{mH}\), and \(C=25 \mu \mathrm{F}\). The circuit is connected to \(10-\mathrm{V}(\mathrm{rms}), 600-\mathrm{Hz} \mathrm{AC}\) source. (a) Is the sum of the voltage drops across \(R, L\), and \(C\) equal to \(10 \mathrm{~V}(\mathrm{rms}) ?\) (b) Which is greatest, the power delivered to the resistor, to the capaci tor, or to the inductor? (c) Find the average power deliv ered to the circuit.

An electromagnetic wave in free space has an electric field of amplitude \(330 \mathrm{~V} / \mathrm{m}\). Find the amplitude of the corresponding magnetic field.

Oxygenated hemoglobin absorbs weakly in the red (hence its red color) and strongly in the near infrared, whereas deoxygenated hemoglobin has the opposite absorption. This fact is used in a "pulse oximeter" to measure oxygen saturation in arterial blood. The device clips onto the end of a person's finger and has two light-emitting diodes \(-\) a red \((660 \mathrm{~nm})\) and an infrared \((940 \mathrm{~nm})-\) and a photocell that detects the amount of light transmitued through the finger at each wavelength. (a) Determine the frequency of each of these light sources. (b) If \(67 \%\) of the energy of the red source is absorbed in the blood, by what factor does the amplitude of the electromagnetic wave change? Hint: The intensity of the wave is equal to the average power per unit area as given by Equation \(21.28\).

A transformer on a pole near a factory steps the voltage down from \(3600 \mathrm{~V}\) (rms) to \(120 \mathrm{~V}\) (rms). The transformer is to deliver \(1000 \mathrm{~kW}\) to the factory at \(90 \%\) efficiency. Find (a) the power delivered to the primary, (b) the current in the primary, and (c) the current in the secondary.

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