/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 cp calt (a) Consider the hydroge... [FREE SOLUTION] | 91Ó°ÊÓ

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cp calt (a) Consider the hydrogen molecule $$\left(\mathrm{H}_{2}\right)$$ to be a simple harmonic oscillator with an equilibrium spacing of \(0.074 \mathrm{nm},\) and estimate the vibrational energy-level spacing for H. The mass of a hydrogen atom is \(1.67 \times 10^{-27} \mathrm{kg}\) . Hint: Estimate the force constant by equating the change in Coulomb repulsion of the protons, when the atoms move slightly closer together than \(r_{0}\) to the "spring" force. That is, assume that the chemical binding force remaing" approximately constant as \(r\) is decreased slightly from \(r_{0}\) ) ( b) Use the results of part (a) to calculate the vibrational energy-level spacing for the deuterium molecule, \(\mathrm{D}_{2}\) . Assume that the spring constant is the same for \(D_{2}\) as for \(\mathrm{H}_{2}\) . The mass of a deuterium atom is \(3.34 \times 10^{-27} \mathrm{kg}\)

Short Answer

Expert verified
For Hâ‚‚: \( \Delta E \approx 7.24 \times 10^{-20} \, \text{J} \). For Dâ‚‚: \( \Delta E_{D} \approx 5.12 \times 10^{-20} \, \text{J} \).

Step by step solution

01

Determine Spring Constant

The force constant \( k \) is determined by equating the change in Coulomb repulsion to the spring force. The Coulomb repulsion between two protons separated by distance \( r \) is given by:\[F_{c} = \frac{k_{e} e^{2}}{r^{2}}\]where \( k_{e} = 8.99 \times 10^{9} \, \text{N m}^{2}/\text{C}^{2} \) is Coulomb's constant, and \( e = 1.6 \times 10^{-19} \, \text{C} \) is the charge of a proton. When atoms move slightly closer from equilibrium separation \( r_{0} = 0.074 \, \text{nm} \), the change in force is equated to the spring force, \( f_{s} = k \cdot \Delta r \). For small displacement, \( \Delta r \), \( k \approx \frac{k_{e} e^{2}}{r_{0}^{2}} \), which calculates \( k \approx 484 \, \text{N/m} \).
02

Calculate Vibrational Energy-Level Spacing for Hâ‚‚

The vibrational energy-level spacing \( \Delta E \) is given by:\[\Delta E = \hbar \sqrt{\frac{k}{\mu}}\]where \( \hbar = 1.05 \times 10^{-34} \, \text{J s} \) is the reduced Planck's constant, and \( \mu \) is the reduced mass of the hydrogen molecule given by \( \mu = \frac{m_{1}m_{2}}{m_{1} + m_{2}} = \frac{(1.67 \times 10^{-27})^{2}}{2 \times 1.67 \times 10^{-27} } \approx 0.835 \times 10^{-27} \, \text{kg} \). Calculate \( \Delta E \):\[\Delta E \approx 7.24 \times 10^{-20} \, \text{J} \].
03

Reduced Mass for Dâ‚‚

For deuterium, the mass is \( m_{D} = 3.34 \times 10^{-27} \, \text{kg} \). The reduced mass \( \mu_{D} \) is calculated similarly:\[\mu_{D} = \frac{(3.34 \times 10^{-27})^{2}}{2 \times 3.34 \times 10^{-27}} = 1.67 \times 10^{-27} \, \text{kg} \].
04

Vibrational Energy-Level Spacing for Dâ‚‚

Using the same spring constant \( k \), calculate the energy-level spacing \( \Delta E_{D} \) for deuterium:\[\Delta E_{D} = \hbar \sqrt{\frac{k}{\mu_{D}}}\]Substitute known values:\[\Delta E_{D} \approx 5.12 \times 10^{-20} \, \text{J} \].

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Key Concepts

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

Vibrational Energy Levels
Vibrational energy levels refer to the discrete energy states that molecules, acting as harmonic oscillators, can occupy due to vibrational motion. Such levels result from quantized energy, meaning molecules can only vibrate at specific energies. This quantization is a fundamental concept of quantum mechanics and influences molecular behavior significantly. In the context of a harmonic oscillator, these energy levels are determined using the formula \( \Delta E = \hbar \sqrt{\frac{k}{\mu}} \), where \( \hbar \) is the reduced Planck's constant, \( k \) is the spring constant, and \( \mu \) is the reduced mass. These energy values affect properties such as molecular spectra, playing a pivotal role in spectroscopic techniques which are used to study molecular compositions. By calculating these levels for hydrogen and deuterium molecules, we understand the quantum mechanical nature of the bonds between atoms in a molecule.
Spring Constant
The spring constant \( k \) represents the stiffness of a bond in a molecular harmonic oscillator, akin to the stiffness of a spring in classical mechanics. It quantifies how much force is needed to stretch or compress the bond by a unit distance. In the context of the hydrogen molecule \( (H_{2}) \), the spring constant is obtained by comparing the spring force with the Coulomb repulsion force when atomic nuclei move slightly closer together from their equilibrium position. Using the formula \( k \approx \frac{k_{e} e^{2}}{r_{0}^{2}} \), where \( k_{e} \) is Coulomb's constant and \( r_{0} \) is the equilibrium separation, we can calculate the spring constant. A large \( k \) means a stiffer bond, requiring more energy to change the bond length, which influences the molecule’s natural vibration frequency.
Reduced Mass
Reduced mass \( \mu \) is a concept used to simplify the motion equations of two interacting bodies, such as atoms in a diatomic molecule like \( H_{2} \). It essentially reduces the two-body problem into a one-body problem. Mathematical manipulations show \( \mu = \frac{m_{1}m_{2}}{m_{1} + m_{2}} \), where \( m_{1} \) and \( m_{2} \) are the masses of the two atoms. Reduced mass influences the vibrational energy levels and dynamics of the molecule significantly. For instance, in this context, calculating \( \mu \) for hydrogen and deuterium shows how atomic mass affects their vibrational behaviors. Heavier molecules like deuterium, with a larger reduced mass, vibrate at lower frequencies compared to lighter molecules such as hydrogen, influencing their corresponding energy-level spacing.
Coulomb Repulsion
Coulomb repulsion describes the electrostatic force experienced when two like-charged particles, such as protons, come close to each other. This principle is crucial for understanding molecular stability and dynamics in chemistry. When atoms within a molecule like \( H_{2} \) or \( D_{2} \) move closer than their equilibrium spacing, the increased repulsion force influences the molecule’s behavior, modulating internal forces such as chemical bonds. This force is calculated using the classic electrostatic formula \( F_{c} = \frac{k_{e} e^{2}}{r^{2}} \), with \( k_{e} \) as Coulomb's constant and \( e \) as the elementary charge. Understanding how this repulsion balances with other forces like the spring-like forces in bonds allows us to estimate parameters such as the spring constant and consequently analyze the vibrational energy levels of these molecular systems.

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Most popular questions from this chapter

CP A variable DC battery is connected in series with a \(125-\Omega\) resistor and a \(p-n\) junction diode that hat hat 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 gap between valence and conduction bands in silicon is 1.12 eV. A nickel nucleus in an excited state emits a gamma-ray photon with wavelength \(9.31 \times 10^{-4}\) nm. How many electrons can be excited from the top of the valence band to the bottom of the conduction band by the absorption of this gamma ray?

An Ionic Bond. (a) Calculate the electric potential energy for a \(\mathrm{K}^{+}\) ion and a Br \(^{-}\) ion separated by a distance of \(0.29 \mathrm{nm},\) the equilibrium separation in the KBr molecule. Treat the ions as point charges. (b) The ionization energy of the potassium atom is 4.3 \(\mathrm{eV}\) . Atomic bromine has an electron affinity of 3.5 eV. Use these data and the results of part (a) to estimate the binding energy of the KBr molecule. Do you expect the actual binding energy to be higher or lower than your estimate? Explain your reasoning.

CALC The one-dimensional calculation of Example 42.4 (Section 42.3\()\) can be extended to three dimensions. For the three-dimensional fce NaCl lattice, the result for the potential energy of a pair of \(\mathrm{Na}^{+}\) and \(C^{-}\) ions due to the electrostatic interaction with all of the ions in the crystal is \(U=-\alpha e^{2} / 4 \pi \epsilon_{0} r,\) where \(\alpha=1.75\) is the Madelung constant. Another contribution to the potential energy is a repulsive interaction at small ionic separation \(r\) due to overlap of the electron clouds. This contribution can be represented by \(A / r^{8},\) where \(A\) is a positive constant, so the expression for the total potential energy is $$U_{\mathrm{tot}}=-\frac{\alpha e^{2}}{4 \pi \epsilon_{0} r}+\frac{A}{r^{8}}$$ (a) Let \(r_{0}\) be the value of the ionic separation \(r\) for which \(U_{\text { ot }}\) is a minimum. Use this definition to find an equation that relates \(r_{0}\) and \(A,\) and use this to write \(U_{\text { ot }}\) in terms of \(r_{0 .}\) For \(\mathrm{NaCl}\) , \(r_{0}=0.281 \mathrm{nm} .\) Obtain a numerical value (in electron volts) of \(U_{\mathrm{tot}}\) for NaCl. (b) The quantity \(-U_{\text { tot }}\) is the energy required to remove a \(\mathrm{Na}^{+}\) ion and a \(\mathrm{Cl}^{-}\) ion from the crystal. Forming a pair of neutral atoms from this pair of ions involves the release of 5.14 eV (the ionization energy of \(\mathrm{Na}\) ) and the expenditure of 3.61 \(\mathrm{eV}\) (the electron affinity of Cl). Use the result of part (a) to calculate the energy required to remove a pair of neutral Na and Cl atoms from the crystal. The experimental value for this quantity is \(6.39 \mathrm{eV} ;\) how well does your calculation agree?

Metallic lithium has a bce crystal structure. Each unit cell is a cube of side length \(a=0.35 \mathrm{nm}\) . (a) For a bce lattice, what is the number of atoms per unit volume? Give your answer in terms of \(a\) . (Hint: How many are there per unit cell?) (b) Use the result of part (a) to calculate the zero- temperature Fermi energy \(E_{\mathrm{PO}}\) for metallic lithium. Assume there is one free electron per atom.

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