Raman Spectrum of Water

Question

This is the Raman spectrum of water. There is $\pu{1635 cm-1}$ Raman peak corresponding to $\ce{HOH}$ and $\pu{3410 cm-1}$ Raman peak corresponding to $\ce{OH}$. But why is there general nonzero background in intensity? What caused it?

This is taken using 532nm wavelength raman system:

This is taken using 785nm wavelength raman system up to 2750 cm-1 only (versus 4000 cm-1 of the first spectrum):

In the above picture. Why is the background dimishing to zero towards 2750 cm-1. My guess is that as wavelength gets higher, water absorbs more. But can it absorb the noise too?

Answer 1

This answer is based on the educated guess that the green excitation Raman spectrum was acquired with a dispersive Raman instrument*.

But why is there general nonzero background in intensity? What caused it?

For any real-life instrumentation, you should never assume to measure 0 signal for 0 analyte concentration (see also Does a calibration curve of absorbance vs concentration need to pass exactly through the origin?) nor expect the baseline to be at zero unless corrected for, even if there really were no Raman scattering outside those two bands, the instrument and/or sample preparation procedures (electronics, optics, masking, cross-sensitivity, ...) will often lead to some non-zero signal.

The CCD in the spectrograph has a so-called dark current. So the raw signal you observe is dark current plus photon signal. Dark current is influenced by settings of the CCD (you trade off sensitivity and dark current) and by temperature.

The constant part of the dark current shouldn't trouble you too much: you can measure and then subtract it (which is in fact what spectrometer software does where you see the baseline at zero). However, as subtracting spectra sums their noise** (error propagation). As Raman measurements typcially suffer from low signal-to-noise-ratio, you may want to avoid this, so Raman spectrometer software usually either outputs raw spectra or offers you the choice whether you want to subtract the dark spectra.

For SNR of Raman spectra, you'll have to take into account shot noise (Photon noise): the noise is directly proportional to the signal intensity, and that includes optical background such as fluorescence or elastically scattered or reflected light as well as your Raman signal. (And then come the other sources of noise such as flicker noise and the various sources of noise related to your application).

For a thorough discussion of the noise properties of Raman spectrometers, there's a whole chaper of McCreery: Raman Spectroscopy for Chemical Analysis, ISBN: 978-0-471-23187-5 on the topic (for FT instruments, see Zhao & McCreery: Multichannel FT-Raman Spectroscopy: Noise Analysis and Performance Assessment, ApplSpec, 1687 - 1697, 1997.)

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About the 785 measurement in comparison:

The green excitation y-axis looks like raw counts (the data itself would reveal more: raw counts are usually integer numbers). In contrast the red spectrum's y-axis goes from 0 to slightly above 1.0. Which means that some transformations have been done (which is impossible to say from the data alone - you'll have to talk to the manufacturer what they do).

 water absorption

Absorbance of in the range below 1100 nm is < 0.2/cm. So yes, you loose about 1/3 of the intensity through 5 mm water and back - but you can avoid that by focusing not as deep into the water.
What hurts your signal more is that the quantum efficiency down there is pretty awful for many CCDs (around 10 %, there's a reason why their spectral range ends in the mid of the -CH stretching vibrations).

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* A rather safe guess, as FT instruments for such short wavelength AFAIK exist at the moment only as "home-built" instruments developed in some research facilities - whoever uses one of those is going to be fully aware that they are using something special.

** more precisely: the variances of independent random variables are additive

Answer 2

A background on Raman:

Samzun (OP), buried in your query might be the taught belief that a Raman scatter spectrum is independent of the excitation wavelength, and this is mostly true, except for a scatter intensity that scales as (1/wavelength)^4 of the excitation. This wavelength-independence makes Raman useful for chemical species identification, and being able to chose the excitation wavelength makes it amenable for various carrier media, including water, for the analyte under observation.

So while the ideal (theoretical) Raman spectrum is fixed, various pieces of your optics will multiply their signature onto the spectrum. As noted in cbeleites' response, a dispersive grating can add its signature. Unlike a simple prism, a grating's signature can be non-monotonic and is often oscillatory, especially in highly efficient holographic blazed gratings.

Another significant contribution is the quantum efficiency of your detector. I suspect you are using a silicon detector for both your 532nm and 785nm excitation. Silicon's quantum efficiency drops off slowly and as it nears 900nm drops off precipitously to zero at 1050nm. At an excitation of 785nm a lot of the scatter will be at those longer wavelengths.

Other pieces of the optics will also add signature to your spectra, e.g. anti-reflection coatings, filters, polarization dependence, etc. A lot of care must be taken to remove these various contributions. As you note, the carrier media (water in your case) can also be an absorber. But in your case for 785nm excitation, you will find that water absorption is not the main contributor, and the detector quantum efficiency is.

If you know the signatures of the various contributors, you can divide them out to get the ideal (wavelength-independent) Raman spectrum of your analyte.

Water spectrum:

An additional question that you ask is about the background around and between the HOH and OH vibrational modes. I do not believe this has been properly addressed in the other replies, and unfortunately the answers are buried in academic literature. This additional background of water is also bothersome for the shot noise is adds across the whole spectrum. Polarization dependent Raman will be the easiest way to convince yourself this broad background in water's spectrum is real.

The likely answer for this broad background is hydrogen-bonding between the water molecules. It loosely bonds multiple water molecules and you then have the vibrational modes of a larger amorphous superstructure. This is very complicated, and so only anecdotal and handwavey discussions are found in literature. This may be unsatisfying, especially to a theoretician, but in your practice it may not matter, e.g. if water's background stays constant and consistent and you know its input variables.

Good luck observing!