How to Read Wavenumber and Convert It into Wavelength

An IR spectrum shows a strong absorption band at \(1700\ \mathrm{cm}^{-1}\). What wavelength of infrared radiation does this represent?

Before calculating, predict whether a band at \(3400\ \mathrm{cm}^{-1}\) would correspond to a longer or shorter wavelength.

The answer is a shorter wavelength. In fact, doubling the wavenumber halves the wavelength. A band at \(1700\ \mathrm{cm}^{-1}\) corresponds to approximately \(5.88\ \mu\mathrm{m}\), while a band at \(3400\ \mathrm{cm}^{-1}\) corresponds to \(2.94\ \mu\mathrm{m}\).

The key relationship is that wavenumber and wavelength are inversely proportional. Once you understand this, you can convert any IR absorption position, interpret broad absorption bands, and avoid some common mistakes when reading spectroscopy data.

01What does wavenumber actually measure?

Wavelength, represented by the Greek letter \(\lambda\), measures the distance between corresponding points on successive waves.

Wavenumber measures something different: how many complete wave cycles fit into a given distance.

Imagine fitting waves into a fixed length of string. If more cycles fit into the same length, each individual wave must be shorter.

This gives us the inverse relationship:

• Higher wavenumber means shorter wavelength.
• Lower wavenumber means longer wavelength.

In IR spectroscopy, wavenumber is normally expressed in reciprocal centimetres, written as \(\mathrm{cm}^{-1}\).

For example, a wavenumber of \(2000\ \mathrm{cm}^{-1}\) means that the corresponding wavelength is \(1/2000\) of a centimetre.

Wavenumber is not quite the same as frequency. Frequency measures the number of wave cycles passing a point each second, whereas wavenumber measures the number of cycles per unit distance. Both are related to wavelength.

The mathematical relationship

The spectroscopic wavenumber, represented by \(\tilde{\nu}\), is the reciprocal of wavelength:

\[ \boxed{\tilde{\nu}=\frac{1}{\lambda}} \]

For this equation:

• \(\tilde{\nu}\) = wavenumber, measured in \(\mathrm{cm}^{-1}\).
• \(\lambda\) = wavelength, measured in centimetres.

The units are important. If the wavenumber is given in reciprocal centimetres, the wavelength calculated directly from the equation will be in centimetres.

You must convert the result if the question asks for micrometres or nanometres.

02How does an IR spectrometer produce a spectrum?

An IR spectrum is not a direct picture of molecular vibrations. It is a graph constructed from measurements of infrared radiation passing through a sample.

Consider a simplified modern Fourier-transform infrared (FTIR) experiment.

The experiment follows this sequence:

1. An infrared source produces radiation across a range of wavelengths.
2. The radiation enters an interferometer, where a beam splitter directs light towards two mirrors. One mirror is fixed, while the other moves.
3. The reflected beams recombine, producing an interference signal.
4. This radiation passes through the sample. Some wavelengths are absorbed by molecular vibrations.
5. A detector measures the transmitted radiation.
6. A computer processes the signal to produce a spectrum showing absorption as a function of wavenumber.

A background measurement is also taken so that the sample’s transmission can be compared with the radiation entering the sample.

The transmittance is calculated using:

\[ T=\frac{I}{I_0} \]

where:

• \(T\) = transmittance (no units).
• \(I\) = transmitted intensity through the sample.
• \(I_0\) = reference intensity measured without the sample.

The instrument can display the resulting information in different ways.

This distinction matters when locating a band.

On a percentage-transmittance spectrum, you find the bottom of the trough. On an absorbance spectrum, you find the top of the peak.

Both represent the same physical absorption. Only the way the measurement is displayed has changed.

03How to convert wavenumber into wavelength

The easiest method is to rearrange the reciprocal relationship and convert directly into micrometres.

Start with:

\[ \tilde{\nu}=\frac{1}{\lambda} \]

Rearrange:

\[ \lambda=\frac{1}{\tilde{\nu}} \]

This gives wavelength in centimetres.

Now remember:

\[ 1\ \mathrm{cm}=10^4\ \mu\mathrm{m} \]

Therefore:

\[ \boxed{\lambda(\mu\mathrm{m})= \frac{10^4}{\tilde{\nu}(\mathrm{cm}^{-1})}} \]

This is the most useful conversion formula for IR spectroscopy.

If the question asks for nanometres instead, use:

\[ \boxed{\lambda(\mathrm{nm})= \frac{10^7}{\tilde{\nu}(\mathrm{cm}^{-1})}} \]

A quick reference table

You can use these values to check whether your calculator result is reasonable.

Notice that the wavelength increases as the wavenumber decreases.

This is a useful way to check your answer before writing it down.

04Reading the spectrum before converting

Knowing the formula is only half the job. You also need to identify the correct wavenumber from the spectrum.

1. Check the horizontal axis

A typical IR spectrum displays wavenumber decreasing from approximately \(4000\ \mathrm{cm}^{-1}\) on the left to \(400\ \mathrm{cm}^{-1}\) on the right.

This is a plotting convention, not a requirement for every spectrum. Always check the actual numbers.

For a conventional decreasing axis:

• \(3500\ \mathrm{cm}^{-1}\) appears towards the left.
• \(1700\ \mathrm{cm}^{-1}\) appears further right.
• \(1000\ \mathrm{cm}^{-1}\) appears even further right.

Since wavelength is inversely proportional to wavenumber, the corresponding wavelengths increase as you move towards the right.

2. Find the correct position of the absorption band

Do not automatically use the left or right edge of a band.

For a relatively narrow absorption, use its central peak position:

• On a %T spectrum, use the bottom of the trough.
• On an absorbance spectrum, use the top of the peak.

If the question provides an exact wavenumber, use that value.

If the band is broad, you may need to convert its entire range rather than selecting one arbitrary position.

3. Convert both ends of a broad band

This is particularly important for broad O-H absorption bands.

Suppose an absorption extends from \(3600\) to \(3200\ \mathrm{cm}^{-1}\).

You should not convert only one endpoint and describe that as the entire absorption.

Instead, convert both values:

\[ \lambda_1=\frac{10^4}{3600} =2.78\ \mu\mathrm{m} \]

\[ \lambda_2=\frac{10^4}{3200} =3.13\ \mu\mathrm{m} \]

Therefore, the absorption extends approximately from:

\[ \boxed{2.78-3.13\ \mu\mathrm{m}} \]

Notice that the order reverses.

The higher wavenumber, \(3600\ \mathrm{cm}^{-1}\), gives the shorter wavelength. The lower wavenumber, \(3200\ \mathrm{cm}^{-1}\), gives the longer wavelength.

If you are given a broad band, report the converted interval clearly.

4. Remember that the conversion is not linear

Here is a less obvious point.

Equal changes in wavenumber do not produce equal changes in wavelength.

For example:

• Moving from \(4000\) to \(3000\ \mathrm{cm}^{-1}\) changes the wavelength from \(2.50\) to \(3.33\ \mu\mathrm{m}\).
• Moving from \(2000\) to \(1000\ \mathrm{cm}^{-1}\) changes the wavelength from \(5.00\) to \(10.0\ \mu\mathrm{m}\).

Both changes involve \(1000\ \mathrm{cm}^{-1}\), but their wavelength differences are very different.

This happens because the relationship involves a reciprocal, not a straight-line conversion.

It also means that a band that looks symmetrical on a wavenumber axis might not look symmetrical when plotted against wavelength.

A reliable conversion method

When given an unfamiliar IR spectrum, use this sequence:

1. Read the horizontal axis and confirm its units.
2. Identify whether the vertical axis is %T or absorbance.
3. Locate the correct absorption position or the boundaries of the band.
4. Record the corresponding wavenumber in \(\mathrm{cm}^{-1}\).
5. Apply \(\lambda=10^4/\tilde{\nu}\) to obtain wavelength in micrometres.
6. If the band is broad, convert both boundaries separately and reverse their order.
7. Check that the final wavelength is consistent with the inverse relationship.

One final chemistry point: converting a wavenumber into wavelength does not provide additional evidence about a molecule. It simply expresses the same absorption using a different unit. When comparing organic structures, you should still consider the position, strength, shape, and combination of absorption bands.

05Worked examples

06Questions and solutions

Question 1

An illustrative percentage-transmittance IR spectrum contains a downward absorption trough at \(2000\ \mathrm{cm}^{-1}\).

Calculate the wavelength of the absorbed infrared radiation in micrometres.

Solution 1

The wavelength is \(5.00\ \mu\mathrm{m}\).

Wavenumber and wavelength are inversely proportional, so we use:

\[ \lambda=\frac{10^4}{\tilde{\nu}} \]

Substitute the given value:

\[ \lambda=\frac{10^4}{2000} \]

\[ \boxed{\lambda=5.00\ \mu\mathrm{m}} \]

The radiation absorbed at \(2000\ \mathrm{cm}^{-1}\) has a wavelength of \(5.00\ \mu\mathrm{m}\).

The fact that the feature is a downward trough tells us that the vertical axis represents transmittance. Its depth does not change the wavelength calculation.

Question 2

An illustrative absorbance spectrum contains a broad upward absorption band extending from \(3600\) to \(3200\ \mathrm{cm}^{-1}\).

The maximum absorbance occurs at \(3400\ \mathrm{cm}^{-1}\).

Calculate:

a) The wavelength corresponding to maximum absorbance.

b) The wavelength range covered by the broad band.

c) Explain why it would be incorrect to report only one wavelength for the entire band.

Solution 2

a) The wavelength at maximum absorbance is \(2.94\ \mu\mathrm{m}\).

Use the reciprocal conversion:

\[ \lambda=\frac{10^4}{3400} \]

\[ \boxed{\lambda=2.94\ \mu\mathrm{m}} \]

b) The broad absorption covers approximately \(2.78-3.13\ \mu\mathrm{m}\).

Convert the higher wavenumber boundary first:

\[ \lambda_1=\frac{10^4}{3600} =2.78\ \mu\mathrm{m} \]

Now convert the lower wavenumber boundary:

\[ \lambda_2=\frac{10^4}{3200} =3.13\ \mu\mathrm{m} \]

Therefore:

\[ \boxed{2.78-3.13\ \mu\mathrm{m}} \]

c) A single wavelength would not describe the entire absorption.

The band extends over a range of wavenumbers, meaning that the sample absorbs radiation across a range of wavelengths.

The value \(2.94\ \mu\mathrm{m}\) identifies the wavelength at maximum absorbance, but it does not describe the boundaries of the entire feature.

This distinction is particularly useful when interpreting broad molecular absorption bands.

Question 3

An illustrative percentage-transmittance spectrum has a horizontal axis that decreases linearly from \(1900\ \mathrm{cm}^{-1}\) at the left-hand end to \(1300\ \mathrm{cm}^{-1}\) at the right-hand end.

The bottom of an absorption trough is located two-thirds of the horizontal distance from the left-hand end.

Calculate:

a) The wavenumber at the bottom of the trough.

b) The corresponding wavelength in micrometres.

Solution 3

a) The trough occurs at \(1500\ \mathrm{cm}^{-1}\).

The total change in wavenumber across the horizontal axis is:

\[ 1900-1300=600\ \mathrm{cm}^{-1} \]

The trough is two-thirds of the distance from the left, so the decrease in wavenumber is:

\[ \frac{2}{3}\times600=400\ \mathrm{cm}^{-1} \]

Therefore:

\[ \tilde{\nu}=1900-400 \]

\[ \boxed{\tilde{\nu}=1500\ \mathrm{cm}^{-1}} \]

b) The corresponding wavelength is \(6.67\ \mu\mathrm{m}\).

Apply the reciprocal conversion:

\[ \lambda=\frac{10^4}{1500} \]

\[ \boxed{\lambda=6.67\ \mu\mathrm{m}} \]

The important detail is that we interpolated using the linear wavenumber scale first.

We did not assume that the same fraction of the horizontal distance would represent the same fraction of a wavelength interval. The reciprocal relationship is non-linear.

Question 4

Two separate absorption bands are observed in an illustrative absorbance spectrum.

Both bands have a full width of exactly \(100\ \mathrm{cm}^{-1}\), measured between their respective baseline crossings.

Their positions are:

• Band A: centred at \(2000\ \mathrm{cm}^{-1}\), extending from \(2050\) to \(1950\ \mathrm{cm}^{-1}\).
• Band B: centred at \(1000\ \mathrm{cm}^{-1}\), extending from \(1050\) to \(950\ \mathrm{cm}^{-1}\).

Calculate the wavelength range of each band and compare their widths in micrometres.

Explain why two bands with identical widths in wavenumber do not have identical widths in wavelength.

Solution 4

Band B is approximately four times wider in wavelength than Band A, despite both having the same width in wavenumber.

We need to convert the boundaries of each band separately.

Band A

The higher wavenumber boundary is \(2050\ \mathrm{cm}^{-1}\):

\[ \lambda_{\text{short}}= \frac{10^4}{2050} =4.88\ \mu\mathrm{m} \]

The lower wavenumber boundary is \(1950\ \mathrm{cm}^{-1}\):

\[ \lambda_{\text{long}}= \frac{10^4}{1950} =5.13\ \mu\mathrm{m} \]

Therefore, Band A covers:

\[ 4.88-5.13\ \mu\mathrm{m} \]

Its wavelength width is:

\[ \Delta\lambda_A=5.13-4.88 \]

\[ \boxed{\Delta\lambda_A=0.25\ \mu\mathrm{m}} \]

Band B

The higher wavenumber boundary is \(1050\ \mathrm{cm}^{-1}\):

\[ \lambda_{\text{short}}= \frac{10^4}{1050} =9.52\ \mu\mathrm{m} \]

The lower wavenumber boundary is \(950\ \mathrm{cm}^{-1}\):

\[ \lambda_{\text{long}}= \frac{10^4}{950} =10.53\ \mu\mathrm{m} \]

Therefore, Band B covers:

\[ 9.52-10.53\ \mu\mathrm{m} \]

Its wavelength width is:

\[ \Delta\lambda_B=10.53-9.52 \]

\[ \boxed{\Delta\lambda_B=1.01\ \mu\mathrm{m}} \]

Comparison

The ratio of the widths is approximately:

\[ \frac{\Delta\lambda_B}{\Delta\lambda_A} = \frac{1.01}{0.25} \]

\[ \boxed{\text{Ratio}\approx4.0} \]

Band B is approximately four times wider in wavelength.

Why?

The reciprocal relationship is non-linear. At lower wavenumbers, a given change in wavenumber produces a much larger change in wavelength.

This is why you must convert both boundaries individually rather than trying to convert a band width using the ordinary reciprocal formula.

Question 5

An IR spectrum contains a single, smooth absorbance band.

The band has the following characteristics on a linear wavenumber axis:

• Maximum absorbance at \(1500\ \mathrm{cm}^{-1}\).
• The band reaches its baseline at \(2000\ \mathrm{cm}^{-1}\) and \(1000\ \mathrm{cm}^{-1}\).
• The band is symmetrical about its maximum on the wavenumber axis.

Two analysts convert the spectrum into wavelength.

Analyst A calculates the wavelength of maximum absorbance using \(1500\ \mathrm{cm}^{-1}\) and obtains \(6.67\ \mu\mathrm{m}\).

Analyst B converts the two band boundaries into wavelengths, obtains \(5.00\ \mu\mathrm{m}\) and \(10.0\ \mu\mathrm{m}\), and reports their midpoint, \(7.50\ \mu\mathrm{m}\), as the wavelength of maximum absorbance.

A third student argues that if the original band is symmetrical, the converted wavelength band must also be symmetrical about its maximum.

Evaluate all three interpretations.

Solution 5

Analyst A is correct: the wavelength of maximum absorbance is \(6.67\ \mu\mathrm{m}\).

Analyst B has correctly calculated the midpoint of the wavelength interval, but incorrectly identified it as the wavelength of maximum absorbance.

The third student’s claim is also incorrect because the reciprocal conversion changes the shape of the band when the horizontal axis is transformed.

Step 1: Calculate the wavelength of maximum absorbance.

The original maximum occurs at:

\[ \tilde{\nu}=1500\ \mathrm{cm}^{-1} \]

Therefore:

\[ \lambda=\frac{10^4}{1500} \]

\[ \boxed{\lambda=6.67\ \mu\mathrm{m}} \]

This is the actual wavelength corresponding to the measured maximum.

Step 2: Convert the two baseline boundaries.

At the higher wavenumber boundary:

\[ \lambda_1=\frac{10^4}{2000} =5.00\ \mu\mathrm{m} \]

At the lower wavenumber boundary:

\[ \lambda_2=\frac{10^4}{1000} =10.0\ \mu\mathrm{m} \]

The complete wavelength interval is therefore:

\[ \boxed{5.00-10.0\ \mu\mathrm{m}} \]

Step 3: Examine Analyst B’s midpoint.

The midpoint of the wavelength boundaries is:

\[ \lambda_{\text{midpoint}} = \frac{5.00+10.0}{2} \]

\[ \lambda_{\text{midpoint}}=7.50\ \mu\mathrm{m} \]

The arithmetic is correct.

However, this is only the midpoint of the wavelength interval. It is not the wavelength of maximum absorbance.

The actual maximum is at \(6.67\ \mu\mathrm{m}\).

Step 4: Explain why the band is no longer symmetrical.

On the original wavenumber axis, the maximum at \(1500\ \mathrm{cm}^{-1}\) is exactly halfway between \(2000\) and \(1000\ \mathrm{cm}^{-1}\).

But after conversion:

• The distance from \(5.00\) to \(6.67\ \mu\mathrm{m}\) is \(1.67\ \mu\mathrm{m}\).
• The distance from \(6.67\) to \(10.0\ \mu\mathrm{m}\) is \(3.33\ \mu\mathrm{m}\).

The maximum is no longer halfway between the boundaries.

The longer-wavelength side spans twice the horizontal distance of the shorter-wavelength side.

This happens because equal changes in wavenumber do not produce equal changes in wavelength.

The chemical absorption has not changed. Only the way its horizontal position is represented has changed.

Final conclusion:

• Analyst A correctly identifies the maximum absorption wavelength as \(6.67\ \mu\mathrm{m}\).
• Analyst B calculates the correct wavelength interval but confuses its midpoint with the absorption maximum.
• The third student’s assumption is incorrect because a reciprocal transformation does not preserve symmetry on a linear horizontal scale.

This is an important distinction when comparing spectra presented using different horizontal axes. You must transform the actual peak position, not simply take the midpoint of the transformed boundaries.

07What to use next

Once you can read wavenumber correctly and convert it into wavelength, the next step is using absorption positions, strengths, and shapes together to interpret an unknown organic compound.

Two useful extensions are:

• Wavelength in HSC Physics: How to Measure It Correctly – revises the physical meaning of wavelength and how it relates to wave behaviour.
• How to Choose Between Organic Structures Using Two IR Features – applies IR absorption positions and band shapes to distinguish possible organic structures.

The main rule to remember is simple: read the wavenumber first, locate the actual absorption feature, and then use the reciprocal relationship to calculate wavelength.