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In IR spectroscopy, the $x$-axis is used to represent wavenumber, in $\mathrm{cm^{-1}}$. Why is wavenumber, equal to $1/\lambda$, used in place of wavelength, which is simply $\lambda$?

Sources I’ve already found explain why it was chosen rather than energy of waves, but the conversion from wavelength to wavenumber is never explained. Below are two relations from Wikipedia, which explain how it can be used in equations, but in all of these cases, $\lambda$ seems to be a choice that’s easier to work with.

What explanations are there, if anything other than “historical reasons”, for why $1/\lambda$ is favored over $\lambda$?

A spectroscopic wavenumber $\tilde\nu$ can be converted into energy per photon $E$ via Planck’s relation:

$$E = hc\tilde\nu$$

It can also be converted into wavelength of light via

$$\lambda = \frac{1}{n\tilde\nu}$$

where $n$ is the refractive index of the medium.

user162485
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sqrtbottle
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    http://en.wikipedia.org/wiki/Wavenumber#In_spectroscopy – Wildcat May 29 '15 at 11:41
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    Wavenumber is directly proportional to energy, so higher wavenumbers correspond to a higher energy by the same factor. (That doesn’t explain why high wavenumbers are usually on the left, though.) That put aside, who still uses IR? – Jan May 29 '15 at 11:51

4 Answers4

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Not only in IR spectroscopy. Wavenumber is unit of energy and therefore you can directly deduce the difference of energy between states.

In addition, humans like to think in acceptably small numbers (0.01 - 10,000). Wavenumber allows this for IR and conveniently supplements the eV unit in small energy separations range. Admittedly, the conversion factor of 8,065.73 won't win beauty contest.

ssavec
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    the same holds for λ as a reflection of energy, though, through E = hc/λ, it's just inverted (though I see your point and think this is a good answer) – sqrtbottle May 29 '15 at 12:38
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    Practically for us chemists the most important conversion factor is 11.96 J per mol = 1 reciprocal centimetre. – J. LS May 29 '15 at 13:41
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    @Sqrtbottle: as an analogy, the temperature is also "wrong", we are accustomed to higher temperature meaning higher numerical value. But for statistical mechanics, you would also like to have temperature expressed in energy units, therefore $\beta = \frac{1}{k_b T}$ – ssavec Jun 01 '15 at 05:53
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The choice to use wavenumbers for infrared spectroscopy (rather than wavelengths, frequencies, or energies) was probably done to provide a range that has both the appearance of width (so that the difference between two peaks is more meaningful) and spans a set of reasonable values that do not contain very large or very small numbers (which are hard to conceptualize). The goal is to be able to easily compare values.

See the following comparison of units/values for the typical range of IR spectroscopy for organic compounds and some "example" values for the absorptions of common bond types:

absorption     cm⁻¹   m        µm     Hz        THz   J          kJ/mol     meV
high end       500    2.00E-5   20     1.5E+13   15    9.94E-21   5.98       62
C-O            1100   9.09E-6   9.09   3.3E+13   33    2.19E-20   13.2       136
C=C            1660   6.02E-6   6.02   5.0E+13   50    3.30E-20   19.9       206
C=O            1720   5.81E-6   5.81   5.2E+13   52    3.42E-20   20.6       213
C-H            3000   3.33E-6   3.33   9.0E+13   90    5.96E-20   35.9       372
O-H            3500   2.86E-6   2.86   1.05E+14  105   6.96E-20   41.9       434
low end        4000   2.50E-6   2.50   1.20E+14  120   7.95E-20   47.9       496

Let's compare especially the peaks for $\ce{C=C}$ and $\ce{C=O}$. These peaks are easily resolvable by all modern FTIR spectrometers, and there is room for peaks to be resolved between them. Only the values in wavenumbers give this sense of resolution intuitively. Modern spectrometers can resolve data to $1.0\ \text{cm}^{-1}$ or better. A resolution of $1.0\ \text{cm}^{-1}$ is equivalent to resolutions of $0.040\ \mu\text{m}$, $0.030\text{ THz}$, $12\text{ kJ/mol}$, and $0.12\text{ meV}$. Which of these resolutions is most intuitive to understand?

Jan
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Ben Norris
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    Well, of course you are falling victim to the exact value logical fallacy. If we were used to kJ/mol it would intuitively immediately make sense to remember that we can resolve peaks around 10 kJ/mol apart. Also, you decidedly say modern spectrometers — but the wavenumber scale was popular way before such a resolution was achieved. Finally, of course we should change NMR scales to something more meaningful as $\pu{0.05ppm}$ makes absolutely no sense */sarcasm* – Jan Oct 20 '17 at 03:52
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Although the question is already a few years old, I hope that I can still add a little perspective to what has already been mentioned by others. First of all, the reason why $1/\lambda$ is preferred over $\lambda$ has been nicely addressed by Ben Norris and ssavec: it provides a scale that is linear in energy so that you can directly compare the distance between lines and also the linewidths. But that being said, why not use other energy units? In fact, you can. You will often find spectra that have frequency units (MHz, GHz, THz, etc) instead of wavenumbers and that is totally fine.

The reason why wavenumber is so popular - besides providing "reasonable" numbers - is partially historical. The first spectroscopic experiments were performed before the advent of quantum mechanics and the relation between the energy and wavelength of light was not known until Planck and Einstein introduced it early in the 20th century. You probably have heard of the Balmer formula (1885) that predicts the emission lines of atomic hydrogen. The original formulation of the Balmer formula is

$$ \lambda = B\left (\frac{n^2}{n^2-2^2} \right ) \text{ with }B=364.50682\,\text{nm}. $$

A few years later (1888) Johannes Rydberg tried to find a similar equation to predict the spectral lines of the alkali atoms (that like hydrogen have a single valence electron). Rydberg did not manage to find an expression similar to that of Balmer. While playing with the data, Rydberg realized that if he used the inverse of the wavelength it was much easier to spot a pattern in the data and he derived the following formula that is now named after him

$$ \frac{1}{\lambda}=R_\text{H}\left (\frac{1}{n_1^2}-\frac{1}{n_2^2} \right ) \text{ with }R_\text{H}=1.09677583\times 10^{-7}\,\text{m}^{-1}. $$

When Rydberg introduced the concept of wavenumber he was unaware of the fact that a wavenumber scale is proportional to an energy scale: he only used it as a mathematical manipulation to find a relation between the spectral lines of alkali atoms. Now that we know that the pattern in a spectrum is easier to interpret if you plot it on an energy scale it does not matter what scale you choose, but many people still prefer the wavenumber scale.

Paul
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Reciprocal length (cm-1) or wavenumber (cm-1) used in vibrational spectroscopy is actually an intuitive concept in reciprocal space (used heavily in diffraction world). From literature: "Reciprocal length is used as a measure of energy. The frequency of a photon yields a certain photon energy, according to the Planck-Einstein relation. Therefore, as reciprocal length is a measure of frequency, it can also be used as a measure of energy. For example, the reciprocal centimetre, cm−1, is an energy unit equaling the energy of a photon with 1 cm wavelength. That energy amounts to approximately 1.24×10−4 eV or 1.986×10−23 J.

The higher the number of inverse length units, the higher the energy.

Community
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Ashraf Khan
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