9.2 Molecular spectra

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By the end of this section, you will be able to:
• Use the concepts of vibrational and rotational energy to describe energy transitions in a diatomic molecule
• Explain key features of a vibrational-rotational energy spectrum of a diatomic molecule
• Estimate allowed energies of a rotating molecule
• Determine the equilibrium separation distance between atoms in a diatomic molecule from the vibrational-rotational absorption spectrum

Molecular energy levels are more complicated than atomic energy levels because molecules can also vibrate and rotate. The energies associated with such motions lie in different ranges and can therefore be studied separately. Electronic transitions are of order 1 eV, vibrational transitions are of order ${10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{eV},$ and rotational transitions are of order ${10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{eV}\text{.}$ For complex molecules, these energy changes are difficult to characterize, so we begin with the simple case of a diatomic molecule.

According to classical mechanics, the energy of rotation of a diatomic molecule is given by

${E}_{r}=\frac{{L}^{2}}{2I},$

where I is the moment of inertia and L is the angular momentum. According to quantum mechanics, the rotational angular momentum is quantized:

$L=\sqrt{l\left(l+1\right)}\hslash \phantom{\rule{0.2em}{0ex}}\left(l=0,1,2,3\text{,...}\right),$

where l is the orbital angular quantum number. The allowed rotational energy level    of a diatomic molecule is therefore

${E}_{r}=l\left(l+1\right)\frac{{\hslash }^{2}}{2I}=l\left(l+1\right){E}_{0r}\phantom{\rule{0.5em}{0ex}}\left(l=0,1,2,3\text{,...}\right),$

where the characteristic rotational energy of a molecule is defined as

${E}_{0r}=\frac{{\hslash }^{2}}{2I}.$

For a diatomic molecule, the moment of inertia with reduced mass $\mu$ is

$I=\mu {r}_{0}^{2},$

where ${r}_{0}$ is the total distance between the atoms. The energy difference between rotational levels is therefore

$\text{Δ}{E}_{r}={E}_{l+1}-{E}_{l}=2\left(l+1\right)\phantom{\rule{0.2em}{0ex}}{E}_{0r}.$

A detailed study of transitions between rotational energy levels brought about by the absorption or emission of radiation (a so-called electric dipole transition    ) requires that

$\text{Δ}l=±1.$

This rule, known as a selection rule    , limits the possible transitions from one quantum state to another. [link] is the selection rule for rotational energy transitions. It applies only to diatomic molecules that have an electric dipole moment. For this reason, symmetric molecules such as ${\text{H}}_{2}$ and ${\text{N}}_{2}$ do not experience rotational energy transitions due to the absorption or emission of electromagnetic radiation.

The rotational energy of hcl

Determine the lowest three rotational energy levels of a hydrogen chloride (HCl) molecule.

Strategy

Hydrogen chloride (HCl) is a diatomic molecule with an equilibrium separation distance of 0.127 nm. Rotational energy levels depend only on the momentum of inertia I and the orbital angular momentum quantum number l (in this case, $l=0$ , 1, and 2). The momentum of inertia depends, in turn, on the equilibrium separation distance (which is given) and the reduced mass, which depends on the masses of the H and Cl atoms.

Solution

First, we compute the reduced mass. If Particle 1 is hydrogen and Particle 2 is chloride, we have

$\mu =\frac{{m}_{1}{m}_{2}}{{m}_{1}+{m}_{2}}=\frac{\left(1.0\phantom{\rule{0.2em}{0ex}}\text{u}\right)\left(35.4\phantom{\rule{0.2em}{0ex}}\text{u}\right)}{1.0\phantom{\rule{0.2em}{0ex}}\text{u}+35.4\phantom{\rule{0.2em}{0ex}}\text{u}}=0.97\phantom{\rule{0.2em}{0ex}}\text{u}=0.97\phantom{\rule{0.2em}{0ex}}\text{u}\left(\frac{931.5\frac{\text{MeV}}{{c}^{2}}}{1\phantom{\rule{0.2em}{0ex}}\text{u}}\right)=906\frac{\text{MeV}}{{c}^{2}}.$

The corresponding rest mass energy is therefore

$\mu {c}^{2}=9.06\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{eV}.$

This allows us to calculate the characteristic energy:

${E}_{0r}=\frac{{\hslash }^{2}}{2I}=\frac{{\hslash }^{2}}{2\left(\mu {r}_{0}^{2}\right)}=\frac{{\left(\hslash c\right)}^{2}}{2\left(\mu {c}^{2}\right){r}_{0}^{2}}=\frac{{\left(197.3\phantom{\rule{0.2em}{0ex}}\text{eV}·\text{nm}\right)}^{2}}{2\left(9.06\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{eV}\right){\left(0.127\phantom{\rule{0.2em}{0ex}}\text{nm}\right)}^{2}}=1.33\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{eV}\text{.}$

(Notice how this expression is written in terms of the rest mass energy. This technique is common in modern physics calculations.) The rotational energy levels are given by

${E}_{r}=l\left(l+1\right)\frac{{\hslash }^{2}}{2I}=l\left(l+1\right){E}_{0r},$

where l is the orbital quantum number. The three lowest rotational energy levels of an HCl molecule are therefore

$l=0;{E}_{r}=0\phantom{\rule{0.2em}{0ex}}\text{eV}\phantom{\rule{0.2em}{0ex}}\left(\text{no rotation}\right),$
$l=1;{E}_{r}=2\phantom{\rule{0.2em}{0ex}}{E}_{0r}=2.66\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{eV},$
$l=2;{E}_{r}=6\phantom{\rule{0.2em}{0ex}}{E}_{0r}=7.99\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{eV}.$

Significance

The rotational spectrum is associated with weak transitions (1/1000 to 1/100 of an eV). By comparison, the energy of an electron in the ground state of hydrogen is $-13.6\phantom{\rule{0.2em}{0ex}}\text{eV}$ .

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