What is a Raman Spectrum | Raman for Beginners | How are Raman Spectra Generated?

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  • @aamirejaz3988
    @aamirejaz3988 2 місяці тому +3

    Your way of explanation is very simple and straightforward... Please make a detailed video on SERS

  • @___Chris___
    @___Chris___ 3 роки тому +2

    I have a question regarding how calibration is achieved: horizontal calibration makes sense, because the wavelength (/wavenumber) of the used laser is known. But what about y-axis calibration? CCDs have a wavelength-dependent sensitivity and a lot of the original intensity of the laser will be lost in the optical system. Is a reference sample the only way to do y-axis calibration or is there another way?

    • @Bruker
      @Bruker  3 роки тому +1

      Dear Christian,
      you are absolutely right and Y-axis must also be calibrated. In our factory, this is done once with a NIST standard before delivery of the Raman microscopes to the customer. This corrects for a CCD's wavelength-sensitivity relation and as this shoud not change over time a user doesnt have to do it again.
      However, if you build your own Raman microscope, you can choose between reference samples or reference lamps to calibrate the Y-axis. It is your choice and both approaches work.
      Please keep in mind, that intensity calibration is not always needed. This error only influences the experiment when you need precise peak height relationship across wavelengths. For example, I want to see the peak A and peak B height ratio with both 532 and 785. Normally they should be same, but with resonance Raman or SERS effect they can change. Conventional Raman measurements, including quantitative maesurements, are NOT influenced by the CCD's wavelength dependency.
      Hope this helps and best regards,
      the Bruker Optics Team

    • @___Chris___
      @___Chris___ 3 роки тому +1

      @@Bruker Great anwer. Thanks for taking the time!

  • @alidermal6126
    @alidermal6126 3 роки тому +2

    شكرا جزيلا

  • @fractalnomics
    @fractalnomics Рік тому

    But the spectra lines are in the infrared (of the EMS) and not the visible (that the laser is) - yes?

  • @chaudry123
    @chaudry123 Рік тому +3

    Wow

  • @shwetasuthar5943
    @shwetasuthar5943 3 роки тому

    Raman shift is Delta E ?

    • @Bruker
      @Bruker  3 роки тому

      Not quite. Delta E is the energy difference, which is linearily related to the wavenumber shift.

  • @ramanwatson1688
    @ramanwatson1688 10 місяців тому +1

    I searched for my name and found this 😅