Pair production and annihilation Unit 1

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  • เผยแพร่เมื่อ 29 ต.ค. 2024

ความคิดเห็น • 56

  • @jacksheldrake7712
    @jacksheldrake7712  9 ปีที่แล้ว +47

    When are people who keep going on about the Lorentz factor going to realise that this is NOT on the AS specification. This in the A2, it is not 'crap', it is actually a teacher who is acutely aware of the qualification that he is teaching. I obviously know these concepts but it is not about me showing off my knowledge, it is about supporting young learners in incremental stages - please don't make arrogant remarks about these videos unless you know the requirements of the course.

    • @Theo.31
      @Theo.31 5 ปีที่แล้ว +2

      jack sheldrake
      You’re video has really helped me to understand this :) I know this was 4 years ago, but I hope the specification hasn’t changed dramatically because this was an old spec. Is it much difference to the latest, 2017 spec?

  • @MrMinevision1
    @MrMinevision1 4 ปีที่แล้ว +4

    My left ear appreciates this video.

  • @Alex-hn7yc
    @Alex-hn7yc 8 ปีที่แล้ว +3

    Literally one of the best online physics channels on youtube. Thanks so much for this :-)

  • @xbreadguy_1080
    @xbreadguy_1080 3 ปีที่แล้ว +3

    My left ear really enjoyed this

  • @jacobmillion7545
    @jacobmillion7545 8 ปีที่แล้ว

    These videos are actual quality. That is why today, you have earned an extra subscriber! Please upload more of this because it really does help! Thanks

  • @athalmangal868
    @athalmangal868 10 ปีที่แล้ว +7

    Thank you so much for this video. It has helped with this part of the topic a lot as i found it difficult to grasp. We need more people like u for every AS subject. Your videos and explanations are excellent. Pls continue make more.

    • @jacksheldrake7712
      @jacksheldrake7712  9 ปีที่แล้ว +1

      Thank you so much for your feedback, I will be making a lot more videos in a couple of months after I have moved house and have my smartboard up again!

  • @godoyricardorg
    @godoyricardorg 9 ปีที่แล้ว +1

    you are really good teaching and great video too, really thanks a lot, it's helping me a lot for a college exam i am going to take on this week

  • @azmathchoudhury8702
    @azmathchoudhury8702 7 ปีที่แล้ว

    scousers have got themselves a great physician! top videos and the examples help so much! thank you

    • @azmathchoudhury8702
      @azmathchoudhury8702 7 ปีที่แล้ว

      as you have worked out the energy and frequency of the electron, in the last example, would this also then by the frequency and energy of the photon then? thanks in advance!

  • @ninisamani
    @ninisamani 10 ปีที่แล้ว

    Thank you so much, Your courses helped me many times, from ALGERIA !!

  • @ben0435
    @ben0435 3 ปีที่แล้ว +1

    my left ear really appreciates this video

  • @gorunmain
    @gorunmain 8 ปีที่แล้ว

    Incredible video, helpful, fast and just brilliant thank you so much.

  • @tomlear5982
    @tomlear5982 8 ปีที่แล้ว

    These videos are so useful!!Keep up the good work.

  • @tanbirpanesar9838
    @tanbirpanesar9838 9 ปีที่แล้ว

    Thanks. Your videos help me a lot in particle physics

  • @jackearl7278
    @jackearl7278 10 ปีที่แล้ว +6

    Should you not multiply your electron volts value by 1.60x10^-13 and not 1.60x10^-19? Because I thought that 1MeV=1.60x10^-13

    • @AbdiiiHD
      @AbdiiiHD 9 ปีที่แล้ว +3

      Jack Earl He converted the 0.511MeV into eV beforehand by timing the 0.511 by 10^6

    • @parasshah524
      @parasshah524 7 ปีที่แล้ว

      doesnt that just make it back intro MeV by doing 10^-19 tho?

    • @shoaibmohammed8461
      @shoaibmohammed8461 7 ปีที่แล้ว +2

      No coz eV into Joules You multiply by (1.6 X 10^-19)
      So he basically saw on his data sheet that mass of an electron is (0.510999 MeV) . he wanted it in eV so did (0.51 X 10^6) to get it from MeV to eV. Then he multiplied (0.51X10^6) by (1.6X10^-19) to get eV into Joules.
      A faster way would be 0.51099 X (1.6 x10^-13) that would give you the same answer.

  • @vin9522
    @vin9522 3 ปีที่แล้ว

    Great video

  • @davidball1222
    @davidball1222 9 ปีที่แล้ว

    Really helpful. I was struggling to understand it but this has helped a lot. thank you. :)

  • @evilstuubi
    @evilstuubi 10 ปีที่แล้ว

    you're a really good teacher, thanks

  • @hasrock36
    @hasrock36 9 ปีที่แล้ว +1

    There appears to be some issues with audio. When using headphones the audio is only playing through one side. I have checked other channels videos and its not a fault with the headphones.

  • @qurbankhan3027
    @qurbankhan3027 4 ปีที่แล้ว

    very helpful video thanks

  • @jamstrr
    @jamstrr 5 ปีที่แล้ว

    Extremely helpful. Thank you!

  • @kevinparamanathan7905
    @kevinparamanathan7905 10 ปีที่แล้ว

    these videos are really helpful for my revision thanks! :D

  • @generalrelativity3490
    @generalrelativity3490 8 ปีที่แล้ว

    Shouldn't the energy considerations in pair production take into account the energies associated with the electromagnetic phenomena that the electron and positron would be associated with?

  • @meharpalbasi4801
    @meharpalbasi4801 8 ปีที่แล้ว

    Will you update your videos for the new spec?

  • @AqibKhan-oi3mp
    @AqibKhan-oi3mp 10 ปีที่แล้ว

    These videos were really helpful thanks! If you could make some for Unit 2 also that'd be great

  • @magyver7890
    @magyver7890 4 ปีที่แล้ว

    Good video

  • @aviavi4684
    @aviavi4684 8 ปีที่แล้ว

    why did you consider mass of electricon as rest mass itself even when it is moving with 0.66C ??

  • @davidc5667
    @davidc5667 10 ปีที่แล้ว

    This was really helpful, thanks.

  • @Nani-ce7tx
    @Nani-ce7tx 9 ปีที่แล้ว

    U ARE AWESOME TEACHERR!

  • @domr8796
    @domr8796 9 ปีที่แล้ว

    thanks, your stuff is so helpful

  • @TS97-a
    @TS97-a 8 ปีที่แล้ว

    Thank you very much! This was really helpful :)

  • @ahmedbyah
    @ahmedbyah 8 ปีที่แล้ว

    Thank you, it's verry helpful

  • @physics-preparation
    @physics-preparation 6 ปีที่แล้ว

    how can i get the smart board which u r using?

  • @jacksheldrake7712
    @jacksheldrake7712  8 ปีที่แล้ว +3

    No, this is AS

  • @DD10Adventures
    @DD10Adventures 5 ปีที่แล้ว

    Saved my life cheers

  • @arifullah7395
    @arifullah7395 2 ปีที่แล้ว

    sir the gamma rays is massless particle than how the law of conservation of momentum hold

    • @karhukivi
      @karhukivi 2 ปีที่แล้ว

      Conservation of energy. The two gamma photons have energies which equate to the kinetic energy of the two particles that are annihilated.

  • @robs5397
    @robs5397 8 ปีที่แล้ว

    these are great

  • @ICEMAN3rdID
    @ICEMAN3rdID 8 ปีที่แล้ว

    Nice.

  • @peterfaigl7876
    @peterfaigl7876 9 ปีที่แล้ว

    Electrons with no kinetic energy? I thought it's not possible to have a stationary electron.

  • @MrMrgodzilla567
    @MrMrgodzilla567 8 ปีที่แล้ว

    Hero

  • @SheepCountings
    @SheepCountings 9 ปีที่แล้ว

    You should consider relativity effects at least, cause the electron is moving really fast. Ekin=(gamma-1)mvv , gamma is the lorentzfactor.

    • @jacksheldrake7712
      @jacksheldrake7712  9 ปีที่แล้ว +6

      Lorentz factor is only considered in A2 turning points, AS students should not concern themselves with this yet.

  • @jas4768
    @jas4768 7 ปีที่แล้ว

    Why are two gamma rays always produced?? Why cant we just have one photon with all of the energy??

    • @shoaibmohammed8461
      @shoaibmohammed8461 7 ปีที่แล้ว +3

      Two photons are released because one photon cannot ensure a total momentum of zero after the collision .

  • @katering2628
    @katering2628 10 ปีที่แล้ว

    THANK YOU

  • @jumpieva
    @jumpieva 3 ปีที่แล้ว

    when you close your eyes i hear gordon ramsey explaining physics

  • @bikkramkesharipradhan2837
    @bikkramkesharipradhan2837 5 ปีที่แล้ว

    Why 2 photons are there

  • @MrJonnyPepper
    @MrJonnyPepper 8 ปีที่แล้ว

    positronic!!!

  • @danielpetka446
    @danielpetka446 6 ปีที่แล้ว +1

    7:05 lol je simply could have divided the number he previously calculated by 2... lmao