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To a good approximation. the hydrogen chloride mole cule, HCl, behaves vibrationally as a quantum harvonie ascillator of spring const ant \(480 \mathrm{~N} / \mathrm{m}\) and with effective oscithating mass just that of the lighter atom, hydrogen If it were in its ground vibtational state, what wave. Iength photon would be just right to bump this motecule up to its next-higher vibrational energy state?

Short Answer

Expert verified
Given the spring constant of the oscillator as \(480 N/m\), the effective mass is the mass of the hydrogen atom (\(1.67 \times 10^{-27} kg\)). Following the steps, we calculate the angular frequency, the frequency of the transition and finally the corresponding wavelength of the photon that helps the molecule transition to the next higher vibrational state.

Step by step solution

01

Formulate Planks-Einstein relation and the energy difference

Start by formulating the Plank-Einstein relation \(E = h \cdot f\) where E is the energy, h is the Plank's constant and f is the frequency. The energy difference between the ground state (n=0) and the first excited state (n=1) of a quantum harmonic oscillator is \(\Delta E = \hbar \cdot \omega\) where \(\hbar\) is reduced Plank's constant and \(\omega\) the angular frequency of the oscillator.
02

Calculate the angular frequency

Next, calculate the angular frequency \(\omega = \sqrt{k/m}\) where k is the spring constant and m is the effective oscillating mass which is the mass of the hydrogen atom. Then calculate \(\hbar \cdot \omega\).
03

Calculate the frequency

According to Planck's equation \(E = h \cdot f\), the frequency of the photon can be obtained by dividing the energy of the transition by Planck's constant, \(f = \Delta E / h\).
04

Calculate the wavelength of the photon

Finally, find the wavelength of the photon using the relation between the speed of light, frequency and wavelength: \(c = \lambda \cdot f\). Therefore, wavelength \(\lambda = c / f\).

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Key Concepts

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

Hydrogen Chloride Molecule
The hydrogen chloride molecule, often referred to as HCl, is a simple diatomic molecule composed of one hydrogen atom and one chlorine atom. Due to this basic structure, it exhibits vibrational and rotational motions that can be analyzed using quantum mechanics. In the context of a quantum harmonic oscillator, which is a model used to describe molecules like HCl, the hydrogen atom vibrates around its equilibrium position with a certain frequency and energy.
The spring constant represents the stiffness of the bond between the hydrogen and chlorine atoms. For HCl, this constant is approximately 480 N/m, indicating a relatively strong chemical bond. It is important when modeling the molecule as a harmonic oscillator because it affects the vibrational frequency of the molecule.
In this model, the mass of the oscillating particle is approximated as the mass of the lighter hydrogen atom, simplifying the calculations and making it easier to analyze the vibrational energy states of the molecule.
Planck-Einstein Relation
The Planck-Einstein relation is a fundamental principle in quantum mechanics that defines the relationship between the energy of a photon and its frequency. This principle can be expressed as:
  • \(E = h \cdot f\)
where \(E\) is the energy, \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J s}\), and \(f\) is the frequency of the photon. This relation is critical for understanding transitions between different energy levels in a quantum system, such as the vibrational states of a molecule.
With regards to HCl and similar molecules, a photon must have the correct energy to induce a transition between vibrational energy states. By applying the Planck-Einstein relation, scientists can calculate the frequency needed for a photon to enable such transitions.
Vibrational Energy States
Molecules like hydrogen chloride have quantized vibrational energy states, which means they can only possess certain discrete energy levels. These states can be likened to the steps on a ladder - the molecule can only "stand" on specific rungs, not in between them. For a quantum harmonic oscillator, these energy levels are determined by the formula:
  • \( E_n = \left( n + \frac{1}{2} \right) \hbar \omega \)
where \(n\) is the vibrational quantum number (n = 0, 1, 2, ...), \(\hbar\) is the reduced Planck's constant, and \(\omega\) is the angular frequency of the oscillator.
The energy difference \(\Delta E\) between two states, such as the ground state \(n=0\) and the first excited state \(n=1\), can be found using \(\Delta E = \hbar \omega\).
  • The ground state (n = 0) has the lowest energy.
  • Transitions occur when the molecule absorbs or emits energy, often in the form of a photon.
Understanding these states is essential for calculating photon energies and predicting molecular behavior.
Photon Wavelength Calculation
In quantum mechanics, the wavelength of a photon associated with a specific energy transition can be calculated using the relationship between its wavelength, frequency, and the speed of light. Once the frequency \(f\) of the photon is determined using the energy difference \(\Delta E\) (from Planck's relation), the wavelength \(\lambda\) can be obtained. This is formulated as:
  • \(c = \lambda \cdot f \)
  • \(\lambda = \frac{c}{f} \)
where \(c\) is the speed of light \(3.00 \times 10^8 \text{ m/s}\). This equation serves to find the specific wavelength of light capable of inducing a vibrational transition.
  • Shorter wavelengths correspond to higher energy photons.
  • Longer wavelengths indicate lower energy photons.
Understanding how to calculate the wavelength not only helps in predicting molecular transitions but also aids in designing experiments and technologies like spectroscopy.

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Most popular questions from this chapter

Exercises \(78-88\) refer to a particle of mass \(m\) described by the wave function $$ \psi(x)=\left\\{\begin{array}{ll} 2 \sqrt{a^{3}} x e^{-a x} & x>0 \\ 0 & x<0 \end{array}\right. $$ Calculate the uncertainty in the particle's momentum.

Exercises \(90-92\) refer to a particle described by the wave function $$ \psi(x)=\sqrt{\frac{2}{\pi}} a^{3 / 2} \frac{1}{x^{2}+a^{2}} $$ (a) Taking the particle's total energy to be 0 , find the potential energy. (b) On the same axes, sketch the wave function and the potential energy. (c) To what region would the particle be restricted classically?

Consider a particle of mass \(m\) and energy \(E\) in a region where the potential energy is a constant \(U_{0}\). greaterthan E. and the region extends to \(x=+\infty\). (a) Guess a physically acceptable solution of the Schrödinger equation in this region and demonstrate that it is a solution. (b) The region noted in part (a) extends from \(x=+1 \mathrm{~nm}\) to \(+\infty\). To the left of \(x=1 \mathrm{~nm}\), the particle's wave function is \(D \cos \left(10^{9} \mathrm{~m}^{-1} x\right)\). Is \(U(x)\) also greater than \(E\) here? (c) The particle's mass \(m\) is \(10^{-30} \mathrm{~kg}\). By how much (in \(\mathrm{eV}\) ) does \(U_{0}\), the potential energy prevailing from \(x=1 \mathrm{~nm}\) to \(+\infty\), exceed the particle's energy?

Exercises \(94-97\) refer to a bound particle of mass \(m\) described by the wave function $$ \psi(x)=A x e^{-x^{2} / 2 b^{2}} $$ For a total energy of 0 , the potential energy is given in Exercise 96. (a) Given these, to what region of the x-axis would a classical particle be restricted? Is the quantum-mechanical particle similarly restricted? (b) Write an expression for the probability that the (quantum-mechanical) particle would be found in the classically forbidden region, leaving it in the form of an integral. (The integral cannot be evaluated in closed form.)

Exercises \(78-88\) refer to a particle of mass \(m\) described by the wave function $$ \psi(x)=\left\\{\begin{array}{ll} 2 \sqrt{a^{3}} x e^{-a x} & x>0 \\ 0 & x<0 \end{array}\right. $$ Verify that the normalization constant \(2 \sqrt{a^{3}}\) is correct.

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