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Spectroscopy: Interaction of light and matter

How UV-Vis and IR radiation can be used to chemical structure and concentrations of solutions

Introduction to spectroscopy

Chemists study how different forms of electromagnetic radiation interact with atoms and molecules. This interaction is known as spectroscopy. Just as there are various types of electromagnetic radiation, there are various types of spectroscopy depending on the frequency of light we are using. We will begin our discussion by considering UV-Vis spectroscopy – that is, what occurs within atoms and molecules when photons in the UV and visible ranges of the spectrum (wavelengths of about 10700 nm) are absorbed or emitted.

UV-Vis spectroscopy

We have mentioned how atoms and molecules can absorb photons, thereby absorbing their energy. Depending on the energy of the photon absorbed or emitted, different phenomena can occur. We will start by considering the simpler case of what happens when a hydrogen atom absorbs light in the visible or UV region of the electromagnetic spectrum.
When an atom absorbs an UV photon or a photon of visible light, the energy of that photon can excite one of that atom’s electrons to a higher energy level. This movement of an electron from a lower energy level to a higher energy level, or from a higher energy back down to a lower energy level, is known as a transition. In order for a transition to occur, the energy of the photon absorbed must be greater than or equal to the difference in energy between the 2 energy levels. However, once the electron is in the excited, higher energy level, it is in a more unstable position than it was when it was in its relaxed, ground state. As such, the electron will quickly fall back down to the lower energy level—and it doing so, it emits a photon with an energy equal to the difference in energy levels. (To help visualize all of this, this video on YouTube provides an excellent example: https://www.youtube.com/watch?v=4jyfi28i928)
The transitions from the higher energy levels down to the second energy level in a hydrogen atom are known as the Balmer series. The greater the distance between energy levels, the higher the frequency of the photon emitted as the electron falls down to the lower energy state.
Excited electrons falling from higher energy levels down to the 2nd energy level in a hydrogen atom will emit photons of different frequencies, and thus different colors of light.
In the diagram above, we have a simplified picture of some of the different energy level transitions possible for our hydrogen atom. Note that the larger the transition between energy levels, the more energy is absorbed/emitted. Therefore, higher frequency photons are associated with the larger energy transitions. For example, when an electron falls from the third energy level to the second energy level, it emits a photon of red light (wavelength of about 700 nm); however, when an electron falls from the sixth energy level to the second energy level (larger transition), it emits a photon of purple light (wavelength of about 400 nm), which is higher in frequency (and thus greater in energy) than red light.
The energy transitions for the electrons of each element are unique, and are distinct from one another. Thus, by examining the colors of light emitted by a particular atom, we can identify that element based upon its emission spectrum. The following shows a few examples of the emission spectra for some common elements:
The atomic emission spectra for H, He, N, O, Ar, Ne, Xe, and Hg.
The atomic emission spectra for various elements. Each thin band in each spectrum corresponds to a single, unique transition between energy levels in an atom. Image from the Rochester Institute of Technology, CC BY-NC-SA 2.0.
Since each emission spectrum is unique to the element, we can think of each of these spectra as being like the “fingerprint” of each element. The thin bands indicate the particular wavelengths of light emitted when electrons in each element fall from an excited state down to a lower energy state. Scientists are able to isolate these different wavelengths by shining the light from excited atoms through a prism, which separates the different wavelengths through the process of refraction. Without a prism, however, we do not see these different wavelengths of light one at a time, but all blended together. Still, the color emitted by each element is quite distinct, which is often useful in a laboratory setting.
In the lab, we can often distinguish between elements by using a flame test. The following picture shows the characteristic green flame that appears when copper metal or copper-containing salts is burned. (Keep in mind that it is the heat energy—a type of electromagnetic radiation—that is able to excite the electrons in each atom.)
A piece of copper metal burns green when exposed to an open flame.
Due to the electronic transitions unique to every copper atom, copper metal burns a characteristic green color when exposed to a flame. Image from Wikipedia, CC BY-SA 3.0.
If we are testing an unknown sample in the lab to determine which elements it contains, we can always use a flame test, and draw a conclusion based on the color of flame that we see. (For more on the use of flame tests, check out this video: https://www.youtube.com/watch?v=9oYF-HxtoYg)

Infrared (IR) spectroscopy: Molecular vibrations

So far, we have been talking about electronic transitions, which occur when photons in the UV-visible range of the spectrum are absorbed by atoms. However, lower energy radiation in the infrared (IR) region of the spectrum can also produce changes within atoms and molecules. This type of radiation is usually not energetic enough to excite electrons, but it will cause the chemical bonds within molecules to vibrate in different ways. Just as the energy required to excite an electron in a particular atom is fixed, the energy required to change the vibration of a particular chemical bond is also fixed. By using special equipment in the lab, chemists can look at the IR absorption spectrum for a particular molecule, and can then use that spectrum to determine what types of chemical bonds are present in the molecule. For example, a chemist might learn from an IR spectrum that a molecule contains carbon-carbon single bonds, carbon-carbon double bonds, carbon-nitrogen single bonds, carbon-oxygen double bonds, to name but a few. Because each of these bonds is different, each will vibrate in a different way, and absorb IR radiation of different wavelengths. Thus, by looking at an IR absorption spectrum , a chemist can make some important determinations about a molecule’s chemical structure.

Spectrophotometry: The Beer-Lambert law

The last type of spectroscopy we will consider is that used to determine the concentration of solutions containing colored compounds. If you’ve ever put food coloring in water, then you already know that the more food coloring you put in, the darker and more colored your solution becomes.
Solutions of potassium permanganate evince a characteristic deep purple color. The greater the concentration of (KMnO4), the darker the solution, and the greater its absorbance.
Solutions of potassium permanganate (KMnO4) of differing concentrations. The more concentrated the solution, the darker the solution becomes, and the greater its absorbance. Image from Flickr, CC BY 2.0.
When a solution becomes darker, it means that it is absorbing more visible light. One of the most commonly used analytical techniques in chemistry is to place a solution of unknown concentration into a spectrophotometer—a device that measures the absorbance of the solution. Absorbance is measured from 0 to 1. An absorbance of zero means that light totally passes through the solution (the solution is completely clear), and an absorbance of 1 means that no light passes through the solution (the solution is completely opaque). Absorbance is related to the concentration of the colored species in solution by the Beer-Lambert law, which is:
Where A is the absorbance (a unitless quantity), ϵ is the molar absorptivity constant (a constant unique to each compound, given in units of M1cm1), l is the path length of the solution container (in cm), and c is the concentration of the solution in molarity (M, or molL).

Example: Using the Beer-Lambert law to find the concentration of a solution

A copper (II) sulfate solution of unknown concentration is placed into a spectrophotometer. A student finds that the solution’s absorbance is 0.462. The molar absorptivity of copper (II) sulfate is 2.81 M1cm1, and the path length of the solution container is 1.00 cm.
What is the concentration of the solution?
First, we can apply the Beer-Lambert Law.
Next, we rearrange the equation to solve for concentration, c.
Lastly, we can plug in our given values and solve for c.
c=0.462(2.81 M1cm1)×(1.00 cm)=0.164 M


Photons carry discrete amounts of energy called quanta which can be transferred to atoms and molecules when photons are absorbed. Depending on the frequency of the electromagnetic radiation, chemists can probe different parts of an atom or molecule's structure using different kinds of spectroscopy. Photons in the UV or visible ranges of the EM spectrum can have sufficient energy to excite electrons. Once those electrons relax back to their ground states, photons will be emitted, and the atom or molecule will give off visible light of specific frequencies. These atomic emission spectra can be used (often informally by using a flame test) to gain insight into the electronic structure and identity of an element.
Atoms and molecules can also absorb and emit lower frequency, IR radiation. IR absorption spectra are useful to chemists because they indicate the chemical structure of a molecule, and the types of bonds it contains. Lastly, spectroscopy can also be used in the laboratory to determine the concentrations of unknown solutions, using the Beer-Lambert Law.

Want to join the conversation?

  • starky tree style avatar for user justformusic2727
    how can a flame test be used to excite electrons when it is stated that IR cant excite electrons but it can only vibrate its chemical bonds?
    (30 votes)
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  • leafers ultimate style avatar for user Shreyas Pai
    In the Beer-Lambert Law section, what is path length?
    (16 votes)
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  • leaf red style avatar for user amaysareen
    Let's assume that the electron of some atom requires x J of energy to move to the next energy level, then can it absorb 2 photons carrying x/2 J of energy to do so? If not, then why and if this can occur, then why don't electrons just absorb energy from more photons of lesser frequency to move to the next energy level?
    (15 votes)
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    • mr pants teal style avatar for user Aaryaman Gupta
      Firstly, it is possible. But it is highly not probable. I mean, the probability of two photons of x/2 J of energy striking the electron at the same time is very very low. And as we already know, electron can't absorb x/2 J first, and wait for another photon of the remaining amount. It either takes all of the energy ( exactly equal to the energy difference of the shells, here as you have taken x) at the same time, or doesn't take at all.
      Hope that helps!
      (22 votes)
  • hopper jumping style avatar for user Sam D
    Above, it says that for a transition to occur the energy of absorbed photon must be equal or greater than the difference in energy between the two levels. But, after a transition the electron quickly falls down to the ground state and in doing so it emits a photon with an energy equal to the difference in energy levels. So if the absorbed photon has a greater energy than the difference between the two energy levels then where does the rest of the energy go after the electron falls back to it's ground state?
    (12 votes)
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  • leaf yellow style avatar for user saransh60
    when we heat the metals, does it takes different amount of heat to excite atoms of different element ?
    (6 votes)
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  • aqualine ultimate style avatar for user Hafsa Kaja Moinudeen
    How can an electron in the hydrogen atom move up to the 6th energy level when excited? The hydrogen atom has only the first energy level right?
    Also, it looks like there are four different electrons that are moving back down to the 2nd energy level, one by emitting a violet photon, the second one by emitting a blue photon, the third by green and the fourth by red; but the hydrogen atom has only one electron, no?
    And, why is the ground state level 2? The only electron of the hydrogen atom resides on the first energy level, Right?

    Thanks very much!
    (7 votes)
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    • mr pants purple style avatar for user Ryan W
      For a hydrogen atom, the lowest energy state has an electron in only the n=1 level, but every atom has access to every energy level. When an atom absorbs a certain amount of energy an electron can be promoted to a higher energy level.
      (4 votes)
  • orange juice squid orange style avatar for user 🌞
    When we stand under the sun, is the heat we feel the result of UV light causing molecular vibrations in our cells, since more kinetic energy through vibrating molecules translates to an increase in temperature? And can the same be said about all heat sources that are also sources of visible/near-visible light, such as ovens and lightbulbs?
    (6 votes)
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    • piceratops seed style avatar for user RogerP
      You are thinking along the right lines, but it is infrared (IR) radiation from the sun, and other sources, that you feel as heat. Although UV light comes from the sun, this is not felt as heat. The UV light, however, can break chemical bonds and this can lead to skin damage or even cancer.
      (6 votes)
  • primosaur seedling style avatar for user Alan Clifford
    In the example emission spectra, what causes the variations in intensity/brightness of the bands? For example, the third band (bright yellow) in the Helium spectrum is surrounded on each side by wide red bands that gradually fade to black. From the explanation, I would have expected very tightly-defined bands of equal brightness.
    (6 votes)
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  • starky seedling style avatar for user Elejune Ace Epanes
    Does the spectra of a element have differences if compared between their different isotopes, like between deutrium and tritium? Do they also have differences in spectra if an atom of a given element is positively or negatively charged?
    (4 votes)
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    • hopper cool style avatar for user Aditya
      hmm, I have NO expertise in this but from what I understand, isotopes shouldn't change the spectrum. Isotopes are the same atom but with different neutrons. The spectrum has to do with electrons and them changing levels. so I think that it would NOT change. Again, I am. NO professional in this subject.
      (3 votes)
  • blobby green style avatar for user 64015197
    The emission spectrum for each element is different, but why exactly is that? There could be different elements with the same number of electrons.
    (3 votes)
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