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Chapter 42: Problem 13

The vibration frequency for the molecule \(\mathrm{HF}\) is \(1.24 \times 10^{14} \mathrm{Hz}\) . The mass of a hydrogen atom is \(1.67 \times 10^{-27} \mathrm{kg}\) , and the mass of a fluorine atom is \(3.15 \times 10^{-26} \mathrm{kg}\) . (a) What is the force constant \(k^{\prime}\) for the interatomic force? (b) What is the spacing between adjacent vibrational energy levels in joules and in electron volts? (c) What is the wavelength of a photon of energy equal to the energy difference between two adjacent vibrational levels? In what region of the spectrum does it lie?

Short Answer

Expert verified
(a) \(k' = 966 \text{ N/m}\); (b) \(\Delta E = 8.213 \times 10^{-20} \text{ J}, 0.513 \text{ eV}\); (c) \(\lambda = 2.42 \times 10^{-6} \text{ m}\), infrared spectrum.

Step by step solution

01

Calculating Reduced Mass

First, calculate the reduced mass (\(\mu\)) of the HF molecule using the formula: \[\mu = \frac{m_1 m_2}{m_1 + m_2}\] where \(m_1 = 1.67 \times 10^{-27} \text{ kg}\) (mass of hydrogen) and \(m_2 = 3.15 \times 10^{-26} \text{ kg}\) (mass of fluorine). This gives: \[\mu = \frac{(1.67 \times 10^{-27})(3.15 \times 10^{-26})}{1.67 \times 10^{-27} + 3.15 \times 10^{-26}} = 1.58 \times 10^{-27} \text{ kg}\]
02

Calculating the Force Constant

The force constant \(k'\) is given by the formula: \[k' = (2 \pi u)^2 \mu\] where \(u = 1.24 \times 10^{14} \text{ Hz}\). Substituting the values in: \[k' = (2 \pi \times 1.24 \times 10^{14})^2 \times 1.58 \times 10^{-27}\] Calculating gives: \[k' = 966 \text{ N/m}\]
03

Energy Spacing Between Vibrational Levels

The spacing between vibrational energy levels \(\Delta E\) is given by: \[\Delta E = h u\] where \(h = 6.626 \times 10^{-34} \text{ Js}\) (Planck's constant). Substituting: \[\Delta E = 6.626 \times 10^{-34} \times 1.24 \times 10^{14} = 8.213 \times 10^{-20} \text{ J}\] To convert this to electron volts (eV), using \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\), gives:\[\Delta E = \frac{8.213 \times 10^{-20}}{1.602 \times 10^{-19}} = 0.513 \text{ eV}\]
04

Finding the Wavelength of the Photon

The wavelength \(\lambda\) is given by \(\lambda = \frac{c}{u}\), where \(c = 3 \times 10^8 \text{ m/s}\) is the speed of light. Substituting gives:\[\lambda = \frac{3 \times 10^8}{1.24 \times 10^{14}} = 2.42 \times 10^{-6} \text{ m}\] This wavelength falls in the infrared region of the electromagnetic spectrum.

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Key Concepts

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

Reduced Mass
The concept of reduced mass plays a significant role in understanding molecular vibrations. In a diatomic molecule like HF, each atom vibrates about their center of mass, and the reduced mass (\( \mu \)) simplifies the calculations of these vibrations. The reduced mass formula is:\[\mu = \frac{m_1 m_2}{m_1 + m_2}\]where \(\, m_1 \) and \(\, m_2 \) are the masses of the hydrogen and fluorine atoms respectively.
The reduced mass effectively combines these two masses into one variable, making it easier to analyze the dynamics of the molecule.
  • Helps in simplifying complex problems involving motion of two masses.
  • Used widely in physics and chemistry for vibrational analysis.
Force Constant
The force constant, denoted as \(\, k' \), measures the stiffness of the bond between two atoms in a molecule. For the HF molecule, it quantifies how much force is needed to change the bond length.
In the context of molecular vibrations, the force constant is calculated using the angular frequency (\(\, u \)) and the reduced mass (\(\, \mu \)): \[k' = (2\, \pi \cdot u)^2 \cdot \mu\]- A higher force constant indicates a stiffer bond.
- Important for predicting vibrational frequencies and energies.
Energy Spacing
Energy spacing (\(\, \Delta E \)) in vibrational levels refers to the difference in energy between consecutive vibrational states. For molecules like HF, these spacings are quantized, meaning they occur at specific energy intervals determined by Planck鈥檚 constant (\(\, h \)) and the frequency (\(\, u \)) of vibration:\[\Delta E = h \cdot u\]
  • Reflects energy absorbed or emitted during transitions between vibrational levels.
  • Key factor in understanding molecular spectroscopy and energy transfer.
Convert to electron volts (eV) to relate to other energy scales in atomic physics.
Photon Wavelength
When a molecule transitions between vibrational levels, it can emit or absorb a photon. The wavelength (\(\, \lambda \)) of this photon is related to the energy difference (\(\, \Delta E \)). Using the speed of light (\(\, c \)), the photon wavelength is given by:\[\lambda = \frac{c}{u}\]- A shorter wavelength corresponds to higher energy transitions.- Calculating this wavelength helps identify which part of the electromagnetic spectrum the transition lies in, such as infrared.
Infrared Spectrum
The infrared spectrum is a part of the electromagnetic spectrum associated with molecular vibrations. Many molecules absorb infrared light during transitions between vibrational levels. The HF molecule's photon wavelength of \(2.42 \times 10^{-6} \text{ m}\) is characteristic of the infrared spectrum.
- Commonly used in spectroscopy to identify molecular bonds and structures.- Infrared absorption corresponds to specific vibrational modes within molecules.- An essential tool for studying molecular dynamics, atmospheric chemistry, and energy transfer processes.

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Most popular questions from this chapter

A variable DC battery is connected in series with a \(125-\Omega\) resistor and a \(p-n\) junction diode that has a saturation current of 0.625 \(\mathrm{mA}\) at room temperature \(\left(20^{\circ} \mathrm{C}\right) .\) When a voltmeter across the \(125-\Omega\) resistor reads 35.0 \(\mathrm{V}\) , what are (a) the voltage across the diode and (b) the resistance of the diode?

The force constant for the internuclear force in a hydrogen molecule \(\left(\mathrm{H}_{2}\right)\) is \(k^{\prime}=576 \mathrm{N} / \mathrm{m} .\) A hydrogen atom has mass \(1.67 \times 10^{-27} \mathrm{kg}\) . Calculate the zero-point vibrational energy for \(\mathrm{H}_{2}\) (that is, the vibrational energy the molecule has in the \(n=0\) ground vibrational level). How does this energy compare in magnitude with the \(\mathrm{H}_{2}\) bond energy of \(-4.48 \mathrm{eV} ?\)

The hydrogen iodide (HI) molecule has equilibrium separation 0.160 \(\mathrm{nm}\) and vibrational frequency \(6.93 \times 10^{13} \mathrm{Hz}\) . The mass of a hydrogen atom is \(1.67 \times 10^{-27} \mathrm{kg}\) , and the mass of an iodine atom is \(2.11 \times 10^{-25} \mathrm{kg}\) . (a) Calculate the moment of inertia of HI about a perpendicular axis through its center of mass. (b) Calculate the wavelength of the photon emitted in each of the following vibration-rotation transitions: (i) \(n=1\) , \(l=1 \rightarrow n=0, \quad l=0 ; \quad(\text { ii) } n=1, \quad l=2 \rightarrow n=0, \quad l=1 ;\) (iii) \(n=2, l=2 \rightarrow n=1, l=3\)

The maximum wavelength of light that a certain silicon photocell can detect is 1.11 \(\mu \mathrm{m}\) . (a) What is the energy gap (in clectron volts) between the valcnce and conduction bands for this photocell? (b) Explain why pure silicon is opaque.

Calculate the wavelengths of (a) a \(6.20-\mathrm{keV} \times\) ray; (b) a 37.6 -eV electron; (c) a 0.0205 -eV neutron.
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