1
$\begingroup$

I know about absorption spectroscopy but really confused between excitation and emission spectroscopy. If they are different, then how?

$\endgroup$
3
$\begingroup$

Absorption:

  1. Light with a specific frequency is selected
  2. This frequency is shined (radiated) onto a molecule (sample)
  3. The amount of light that shines through the molecule is measured, ie. the detector is behind the sample (180 degrees).
  4. Step 1 is repeated with a different frequency.

When the sample absorbs a specific frequency, less light of that frequency will shine through to the detector. The machine records this and a peak appears in the spectrum for that frequency.

Frequency is usually measured in wavelength, and a wide range of wavelengths are applicable. In UV-VIS, this can be eg. 300-800 nm, so 800 nm is measured, then 799 nm, then 798 nm...down to 300 nm. In IR, you will go into micrometers. In X-ray, you will be around 10 nm I think.

Emission:

  1. A light with a specific frequency is selected
  2. The sample is irradiated with this frequency.
  3. The amount of light that shines from the molecule is measured, ie. the detector is to the side of the sample (90 degrees).
  4. The emitted light is measured for a range of frequencies
  5. Step 1 is repeated with a different frequency.

In both processes, the sample absorbs light. The major difference here is the detector being located to the side of the sample. This allows it to ignore the light being shined onto the molecule, and instead measure light that is emitted from the sample. Think of the incident light as a column. In absorption spectroscopy, that column is absorbed by the sample and continues to the other side, where the detector can measure it. In emission spectroscopy, you want to avoid the column of light. Think of the sample like a light bulb that shines light in all directions. The detector in fluorescence is placed to the side, avoiding the column of light and picking up the light from the sample instead. The advantage to fluorescence is you can often selectively measure a moiety (part of the molecule) with superior sensitivity, whereas in absorption impurities or stray light are more prone to interfere with the signal and also leads to higher concentrations usually being necessary.

The reason the emission wavelength detected is different than the frequency used to irradiate it is due to Stoke's Shifts.

Stoke's Shift

Stoke's Shift requires basic knowledge of the difference between electronic and vibrational states. A molecule has electronic states, S=0, 1, 2, 3...etc. Within each electronic state is a set of vibrational states, v=1,2,3.... For example, two electrons can be in the 0 electronic state (termed ground state), but one electron can be in the v=2 vibrational state, while the other electron in v=4.

This is easier to picture if you look up Stoke's Shift, or specifically the diagram on Wikipedia's page for Fluorescence.

For example, in emission spectroscopy, an electron absorbs light, being excited electronically from S=0, v=1, to S=1, v=2. Vibrational relaxations will occur before electronic relaxations. That means the electron will first relax from S=1, v=2 to S=1, v=0. After this, it relaxes to S=0, emitting a photon in the process for detection. Because it relaxed its vibrational state first, energy was lost not in the form of light (thermodynamically). As a result, the photon finally emitted is at a lower energy than the photon that had been absorbed. That's the rudimentary explanation.

$\endgroup$
  • $\begingroup$ Thanks!! But is the position of detector makes all the difference between them? can we still calculate the emission from the sample if we keep the detector just behind the sample? $\endgroup$ – daredvill Feb 23 '16 at 6:03
  • $\begingroup$ No, for 2 reasons: 1. The emission wavelength is not known. 2. The incident light (the light used to excite the sample) shines like a column through and behind the sample. The first reason means the detector has to scan every possible wavelength to test for the emission wavelength. This is a problem for the second reason. While scanning every wavelength, it will also be detecting the massive signal from the incident wavelength which will not only overlap with the emission signal, but it will drown it out with its intensity. You need to avoid detecting the incident light, hence 90 degrees. $\endgroup$ – Blaise Feb 24 '16 at 16:49
1
$\begingroup$

In order to appreciate the difference between UV-VIS emission and excitation spectroscopy it is crucial to have a closer look at the instrument that is used to carry out both techniques, i.e. the Fluorometer.

The fundamental optical components of a Fluorometer are: a lamp that provides the light excitation, a monochromator of excitation (Mex) to select a monochromatic radiation (the radiation has always a certain bandwidth so it is never monochromatic) used to excite the sample, a monochromator of emission (Mem) that selects the radiadion coming out from the sample, this can be luminescence (fluorescence or phosphorescence), raman or scattered light (with a 90 degree geometry used in the experiment the scattered light is greatly reduced but not completely eliminated). Finally a detector, usually a photomultiplier.

In EMISSION spectroscopy the Mex is fixed at the wavelenght of excitation and the Mem scans, i.e. it lets pass different monochromatic radiation sequentially, in a selected wavelenght range. The final spectrum (emission) results in emission-intensity vs wavelength.

In EXCITATION spectroscopy the Mex scans in a wavelength range equal to that used in absorption spectroscopy. The Mem is fixed at a certain wavelength. What you want to have is a spectrum (excitation) that is proportional to the absorption spectrum. If that is the case it means that all the excited states (those that have been populated by exciting the sample in a Mex scanning mode) contribute to populate the emitting state (the one that is monitored by fixing the Mem). If that is not the case it means that not all the excited states contribute to populate the emitting state, i.e. there are other deactivation paths for the excited states located at higher energy compared to the emitting state.

In other words, with emission spectroscopy you have information on the capability of your sample to emit light. On the other hand, with the excitation spectroscopy you want to investigate whether high-energy excited states communicate with the emitting state, i.e. if their population deactivate towards the emitting state or not. This also implies that to perform an excitation spectrum you have to have an emitting compound. Thus, first you run an emission experiment and then an excitation.

$\endgroup$
1
$\begingroup$

There are three types of spectra here, absorption, emission (fluorescence, phosphorescence, in general 'luminescence') and excitation.
(a) Absorption as you know is a transition from the ground state to the first, second or higher electronically excited states. These states will decay back to the ground state, assuming no chemistry from an excited state occurs, by radiationless and radiative transitions.
Radiationless transitions occur from a singlet excited state to the ground state (internal conversion) or to a triplet state (intersystem crossing).
(b) In competition with these transitions is fluorescence from the singlet and phosphorescence from the triplet to ground state. This is the second type of spectrum and is obtained by fixing the excitation wavelength and scanning the emission monochromator over different wavelengths and recording the signal vs. emission wavelength.
(c) The third type of spectrum is excitation. To obtain this the detector is set at the required wavelength in the emission spectrum and the excitation wavelength is scanned and the signal recorded vs. excitation wavelength. This spectrum is related to the absorption spectrum, but differs from it in intensity because the Franck-Condon factors are different in absorption and emission. Additionally, suppose that radiationless transitions are greater from one excited state than another then less emission will be observed, compared to that expected from the absorption spectrum, and so a weak excitation spectrum will now be observed at these wavelengths. Thus the excitation spectrum is like an 'action' spectrum. In general excitation spectra are difficult to interpret accurately.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.