Vibrational plus Rotational Energy of Molecules

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  • Опубликовано: 3 дек 2024

Комментарии • 20

  • @carl-henrikhanquist9956
    @carl-henrikhanquist9956 8 лет назад +1

    Thanks, helped me understand how the formulas are used together.

  • @thejamesmom
    @thejamesmom 3 года назад +3

    Finally a video without a heavy Indian accent

    • @mr.ahmeed
      @mr.ahmeed 25 дней назад

      خخخخخخ خرب 🤣🤣🤣

  • @ClouduSan
    @ClouduSan 4 года назад

    Damnnn this is before the super high quality newer videos, what a difference haha :)

  • @hansfeldmeier6267
    @hansfeldmeier6267 5 лет назад +7

    Unfortunately, there is a substantial mistake in this video. In the text and also in the explanations of the speaker it is said that an electron makes a transition when changing the rotational quantum number l or vibrational quantum number v. It is the whole molecule including the nuclei which changes its state. Somewhat simplifying one may say that for larger l the nuclei rotate faster and drag the electrons with them. For larger v the nuclei oscillate back and forth with larger amplitude and drag the electron with them.
    The energies needed to make a transition from one electron orbit to another one are much higher.

    • @kelseyadler5812
      @kelseyadler5812 Год назад

      Thank you so much for taking the time to write this out. I was a bit skeptical/confused by that.

    • @rav2n
      @rav2n 2 месяца назад

      Can you explain it better? Maybe an audio recording with a piece of paper having diagrams and equations uploaded to you channel and a link shared here?

  • @IEleMenTIx
    @IEleMenTIx 7 лет назад +1

    One thing is still unclear to me. Do atoms vibrate faster (frequency) or stronger (amplitude) about their equilibrium position when you heat a crystalline solid?
    By your equation it looks like if only the frequency would change. I have thought about it and what do you think about this:
    So the kinetic energy distribution is described most simply by = 1/2 m* where the average squared velocity can be described as v^2=omega^2*x^2 so x would be the displacement of the vibrating atoms and omega the angular velocity. Again these terms depend on the spring constant k and the mass m -> omega^2= k/m... if you look at an anharmonic potential curve this parabolic approximation would only describe the lowest part of the potential energy curve. Due to the anharmonic character the spring constant has to change change, therefore the frequency f has to change by f=1/2pi * sqrt(k/m)
    So what do you think?

  • @Brolnox
    @Brolnox 7 лет назад

    Very clear explanation. Well done.

  • @benbedraabdssamad1413
    @benbedraabdssamad1413 6 лет назад +1

    Thank so much! Hmmm, I think that electrons have nothing to do with vibrational and rotaional energies! the emission spectrum of vibration/rotation of molecules is produced by nuclear transitions between their energy levels (quantum states) and not by electonic transitions.

  • @jesusibanezsagasti9138
    @jesusibanezsagasti9138 8 лет назад

    Very well explained, thank you.

  • @rav2n
    @rav2n 2 месяца назад

    Do these rotational and vibrational effects exists in solid state and also cause electrons to transition to other levels?

  • @chemistrybysaadanwar
    @chemistrybysaadanwar 4 года назад

    Well explained

  • @brothersffgameplay77
    @brothersffgameplay77 6 лет назад

    Very nice....thank you so much

  • @vatalay8899
    @vatalay8899 5 лет назад

    Legend!

  • @aliyajamal3757
    @aliyajamal3757 7 лет назад

    well explanation

  • @mohamedmhamedezzine8908
    @mohamedmhamedezzine8908 5 лет назад

    Thanks

  • @somethingfishy1088
    @somethingfishy1088 3 года назад

    thanks bro

  • @sanimmid
    @sanimmid 7 лет назад

    So easy bro

  • @nomad3976
    @nomad3976 4 года назад

    Great Lesson.
    Terrible microphone