# 7.5 The quantum harmonic oscillator  (Page 3/4)

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## Vibrational energies of the hydrogen chloride molecule

The HCl diatomic molecule consists of one chlorine atom and one hydrogen atom. Because the chlorine atom is 35 times more massive than the hydrogen atom, the vibrations of the HCl molecule can be quite well approximated by assuming that the Cl atom is motionless and the H atom performs harmonic oscillations due to an elastic molecular force modeled by Hooke’s law. The infrared vibrational spectrum measured for hydrogen chloride has the lowest-frequency line centered at $f=8.88\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{ }13}\phantom{\rule{0.2em}{0ex}}\text{Hz}$ . What is the spacing between the vibrational energies of this molecule? What is the force constant k of the atomic bond in the HCl molecule?

## Strategy

The lowest-frequency line corresponds to the emission of lowest-frequency photons. These photons are emitted when the molecule makes a transition between two adjacent vibrational energy levels. Assuming that energy levels are equally spaced, we use [link] to estimate the spacing. The molecule is well approximated by treating the Cl atom as being infinitely heavy and the H atom as the mass m that performs the oscillations. Treating this molecular system as a classical oscillator, the force constant is found from the classical relation $k=m\text{ }{\omega }^{\text{ }2}$ .

## Solution

The energy spacing is

$\text{Δ}E=h\text{ }f=\left(4.14\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{ }-15}\text{eV}·\text{s}\right)\left(8.88\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{ }13}\phantom{\rule{0.2em}{0ex}}\text{Hz}\right)=0.368\phantom{\rule{0.2em}{0ex}}\text{eV}\text{.}$

The force constant is

$k=m\text{ }{\omega }^{\text{ }2}=m\text{ }{\left(2\pi f\right)}^{2}=\left(1.67\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{ }-27}\phantom{\rule{0.2em}{0ex}}\text{kg}\right){\left(2\pi \phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}8.88\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{ }13}\phantom{\rule{0.2em}{0ex}}\text{Hz}\right)}^{2}=520\phantom{\rule{0.2em}{0ex}}\text{N}\text{/}\text{m}.$

## Significance

The force between atoms in an HCl molecule is surprisingly strong. The typical energy released in energy transitions between vibrational levels is in the infrared range. As we will see later, transitions in between vibrational energy levels of a diatomic molecule often accompany transitions between rotational energy levels.

Check Your Understanding The vibrational frequency of the hydrogen iodide HI diatomic molecule is $6.69\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{ }13}\phantom{\rule{0.2em}{0ex}}\text{Hz}$ . (a) What is the force constant of the molecular bond between the hydrogen and the iodine atoms? (b) What is the energy of the emitted photon when this molecule makes a transition between adjacent vibrational energy levels?

a. 295 N/m; b. 0.277 eV

The quantum oscillator differs from the classic oscillator in three ways:

First, the ground state of a quantum oscillator is ${E}_{0}=\hslash \omega \text{/}2,$ not zero. In the classical view, the lowest energy is zero. The nonexistence of a zero-energy state is common for all quantum-mechanical systems because of omnipresent fluctuations that are a consequence of the Heisenberg uncertainty principle. If a quantum particle sat motionless at the bottom of the potential well, its momentum as well as its position would have to be simultaneously exact, which would violate the Heisenberg uncertainty principle. Therefore, the lowest-energy state must be characterized by uncertainties in momentum and in position, so the ground state of a quantum particle must lie above the bottom of the potential well.

Second, a particle in a quantum harmonic oscillator potential can be found with nonzero probability outside the interval $\text{−}A\le x\le +A$ . In a classic formulation of the problem, the particle would not have any energy to be in this region. The probability of finding a ground-state quantum particle in the classically forbidden region is about 16%.

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No
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Not so more
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DHEERAJ
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But ask anything changes itself with respect to time or surrounding A Not any harmful radiation
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i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
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there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
Prema
this msg is out of mistake. sorry friends​.
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