Blackbody Curve and Wien's Law Explained (UPDATED)

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  • Опубліковано 13 гру 2024

КОМЕНТАРІ • 40

  • @CharlesLeTower
    @CharlesLeTower 4 роки тому +26

    I’m 36 yrs old, came here out of curiosity and I can’t believe they teach this in high school nowadays.
    Nice vid.

    • @PhysicsHigh
      @PhysicsHigh  4 роки тому +2

      Thanks. Sure do. Modern physics and the lead up to, is an important topic nowadays.

    • @577niccy
      @577niccy 3 роки тому +2

      i agree

  • @douglasstrother6584
    @douglasstrother6584 6 місяців тому +1

    Planck's approach was to analyze the entropy of blackbody radiation as a function of energy. To make both high-frequency and low-frequency data consistent with the Second Law of Thermodynamics, he included an additional "guess" term proportional to the frequency (hf); this results in Planck's Law. Planck's application of Boltzmann's Statistical Mechanics to justify his guess then led to his revolutionary conclusion that the material of the walls emit and absorb radiation in discrete quanta.
    A paper titled "Planck’s Route to the Black Body Radiation Formula and Quantization" by Michael Fowler (7/25/08) gives a nice discussion.

  • @caklnl
    @caklnl 3 роки тому +6

    At the 5:24 mark there’s a conflation between spectral intensity (radiance) and the physicist’s definition of intensity being the power arriving at a unit area some distance from the source. The simulator is labeled “Intensity” which leads to the confusion. When you set the simulator to the Sun’s temperature of 5750 K (it’s actually 5770 but you have to go with the closest setting available) you get 6.20 x 10^7 W/m2. That’s exactly what you get if you divide the Sun’s luminosity by the area of the photosphere. So this “Intensity” is actually the Radiance or Spectral Intensity. Shouldn’t be referring to the Inverse Square Law because that applies to the physicist’s intensity. The blackbody curve should be the same for a 3000 K red dwarf and a 3000 K red giant. It is if we’re considering radiance (how much power it’s emitting through each unit area of its surface.). It would not be the same curve if we were measuring power arriving at a unit area some distance from the star. They would peak at same wavelength but the red giant’s curve would have much more area since it is far more luminous.

    • @yugzed2233
      @yugzed2233 3 роки тому +1

      This is what I am confused with, too.

  • @rayirthsivakumar8202
    @rayirthsivakumar8202 3 роки тому +8

    this was one of the best explanations i have ever heard. Thank you sir

  • @arghojitsaha3568
    @arghojitsaha3568 2 роки тому +1

    Thank you sir...im a school student of class12,from India...this video was helpful!

  • @juliagli1973
    @juliagli1973 4 роки тому +4

    Thank you, very helpful for university

  • @tristanchristensen5963
    @tristanchristensen5963 3 роки тому

    This was the best explanation of Wien's Law I've seen. Whatever that simulation was is really helpful

  • @4pharaoh
    @4pharaoh Рік тому

    Since the wavelength is the inverse of frequency times the speed of light isn’t another relationship simply:
    The peak temp T=6.6 *F where F is the Peak frequency in THz (THz =1e12 Hz)

  • @manuellorena1322
    @manuellorena1322 5 місяців тому

    Excelent video! Thanks

  • @djtailik3761
    @djtailik3761 4 роки тому +7

    Why are you sitting in front of the Y-Axis?

  • @alarialessard-northrup8834
    @alarialessard-northrup8834 3 роки тому +1

    this was so helpful thank you

  • @saveearth9816
    @saveearth9816 3 місяці тому

    ... I had watched many videos about this phenomenon (BLACK BODY RADIATION)
    what was confusing me is how they construct the curve for each certain temperature.... Then I understand that they heat the matter for e. g to 3000 K the body will glow (emit visible light +IR&UV ray) then they will pass this light through glassy prism which will analyse it into the spectrum including visible colors +IR&UV... Then they measure the intensity of each part of the spectrum (starting from IR...then visible light colors ended by UV)
    Plot wave length (horizontal) & intensity longitudinal
    THANKS & SORRY FOR THE BAD ENGLISH I USED

  • @lilcookie7118
    @lilcookie7118 3 роки тому +3

    I really wish you help me in this ! 2:03 I watched Planck's video but can't quite understand the concept. Why is the ultra violet intensity ( radiation ) very low and another question why is their a peak in the graph. Thank U a lot for all the hard work you put in !

    • @shoutitallloud
      @shoutitallloud Рік тому +1

      If you have to carry away some amount of energy, you need more low capacity carriers, but less high capacity carriers. UV waves are like high capacity, and IR are low capacity. And there's one rule that you can't use any type of carrier until it's fully filled. (that relates to quantum nature) So if you total energy is small (low temperature), you can't use half-filled high capacity UV waves. You would emmit energy in small portions with low capacity IR waves. But if temperature is high you would rather prefer to use high capacity UV waves.

    • @lilcookie7118
      @lilcookie7118 Рік тому +1

      @@shoutitallloud So, the ultraviolet radiation has high capacity carriers that has more energy thus less intensity but the IR has low capacity carriers thus more carriers - radiation -, high intensity and less energy with each individual photon. Am I correct?

    • @shoutitallloud
      @shoutitallloud Рік тому

      @@lilcookie7118 If your reasoning comes to conlusion that I understand as "IR photons have low energy, and UV photons have high energy" - you are correct.

  • @enochbrown8178
    @enochbrown8178 Рік тому

    Simply wonderful. Thank you!!!

  • @billygraham5589
    @billygraham5589 2 роки тому

    A question:
    Is the total energy represented by the area under the intensity curve to the LEFT of the peak intensity point EQUAL TO the total energy to the right of it?

  • @johnjeremias9437
    @johnjeremias9437 4 роки тому +1

    thanks, very cleay explanation, enjoyed your presentation

  • @samh-l5711
    @samh-l5711 4 роки тому +1

    Thank you! Very well made and easy to understand. :)

  • @kniridgaming1005
    @kniridgaming1005 3 роки тому

    loved the video

  • @ZiadHelou
    @ZiadHelou 3 роки тому

    Finally understood..Thank you

  • @soweitwegvonperfection7150
    @soweitwegvonperfection7150 4 роки тому +1

    Thank you Paul :)

    • @PhysicsHigh
      @PhysicsHigh  4 роки тому

      You’re welcome. Glad you like it.

  • @Buddha-Einstein
    @Buddha-Einstein 9 місяців тому

    Purely INTUITIVELY SPEAKING, HOW DOES QUANTIZATION show intensity going down at higher frequencies? Put the math aside, what stops energy from taking next quanta of hf?

  • @sollinw
    @sollinw 2 роки тому

    Very good!

  • @richardoberhammer1730
    @richardoberhammer1730 4 роки тому +3

    In the first minute you say that a black body absorbs all radiation that falls on it and then re-emits electromagnetic radiation, and a star is a generator of light with no reflection going on. I don’t get the connection between the two. Why are you talking about a star as if it has a relation to a black body?

    • @PhysicsHigh
      @PhysicsHigh  4 роки тому +1

      A star is the closest thing to a black body. True, it’s not absorbing energy externally but it is internally through fusion. It’s emission curve is a black body curve. Maybe if I say that a black body only radiates energy through emission and there’s no reflection.

  • @SUCKYPICK
    @SUCKYPICK 2 роки тому

    What program is this you are using?

    • @SUCKYPICK
      @SUCKYPICK 2 роки тому

      Never mind, thanks google! The interactive scale he is using is here: phet.colorado.edu/sims/html/blackbody-spectrum/latest/blackbody-spectrum_en.html

  • @hollywoodbanayad7755
    @hollywoodbanayad7755 4 роки тому +1

    Wein's constant is more roughly equal to 2.898e-3, but you still got all of your points across to me. Thank you!!

  • @ajaykumar-im9hb
    @ajaykumar-im9hb 2 роки тому

    Nobody explains why intensity is less in UV area. The lecture could be made more simple and to the point.

    • @shoutitallloud
      @shoutitallloud Рік тому

      If you have to carry away some amount of energy, you need more low capacity carriers, but less high capacity carriers. UV waves are like high capacity, and IR are low capacity. And there's one rule that you can't use any type of carrier until it's fully filled. (that relates to quantum nature) So if you total energy is small (low temperature), you can't use half-filled high capacity UV waves. You would emmit energy in small portions with low capacity IR waves. But if temperature is high you would rather prefer to use high capacity UV waves.

  • @fanR6580
    @fanR6580 4 роки тому

    Ojalá alguien hiciera un video así en español :c

  • @eatyourvegetables226
    @eatyourvegetables226 2 роки тому

    What program are you using?