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Physics: Post your doubts here!

Discussion in 'International A And AS Level' started by XPFMember, Sep 29, 2011.

  1. Deeksha Tamang

    Deeksha Tamang

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    thank you so much
     
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  3. anastasia grey113

    anastasia grey113

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    yh shes right ive left these ques
     
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  4. Hussein Yasir

    Hussein Yasir

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    T
    Lots of thanks:)
     
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  5. Holmes

    Holmes

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  6. Deeksha Tamang

    Deeksha Tamang

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  7. GarryTheGhost

    GarryTheGhost

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    can someone help me for part 1
    i dont understand it, i tried comparing the question to the observation of photoelectric effect but still got no luck :c
    for the second part:
    it says that wavelength is reduced so i thought that frequency is increased now since the wavelength already exceeds the threshold frequency it means that all it will do is allow electrons to have a greater maximum kinetic energy. But the markscheme says photon energy is larger so hence larger to kenietic energy is also larger what does that even mean!!!!!
    and more maximum photoelectric current it says fewer photons per unit time so maximum current is smaller????
     

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  8. adwait Chooromoney

    adwait Chooromoney

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    Explain the function of the non-uniform magnetic field that is superimposed on a large uniform
    magnetic field in diagnosis using nuclear magnetic resonance imaging (NMRI) (4marks) nov16/p42

    upload_2018-4-15_19-10-15.png
    I am having difficulty to write a model answer for this question.
     
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  9. anastasia grey113

    anastasia grey113

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    When nuclei are placed in magnetic field, they precess at Larmor frequency and emit r.f. pulses. This frequency depends on strength of magnetic field as w = Bγ. Different rf pulse frequency emitted allow nuclei to be located. Changing field also allows position of detection to be changed.
     
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  10. anastasia grey113

    anastasia grey113

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    For part 1, the photoelectric effect will not help because it supports the particulate nature of light.
    Look into the wave nature instead which is very different.
    According to that theory, there is SUFFICIENT TIME DELAY between electron emmision and light incidence.
    This is because electrons GAIN ENERGY FROM MORE THAN ONE PHOTONS and when they 'collect' all this energy and it's equal to work function, they escape. This obvio takes some time.
    This also means, that if suppose a low frequency light is shone, the electrons will still be emitted, only more time will be taken because as low energy photons are present it will take more time for electrons to gather up energy equal to work function. So electrons will be EMITTED AT ALL FREQUENCIES.
    But if intensity is increased, more photons given out per unit time so electrons will gain more energy and more quickly so their max. kinetic energy will also increase.
    Write the points I wrote in capital.
    And remember: THIS IS JUST THE WAVE THEORY OF LIGHT. THE PARTICULATE THEORY IS VERY DIFFERENT FROM THIS.

    Second part
    Remember this:
    1- When wavelength decreases, frequency increases.
    2- When frequency increases, photon energy increases.
    3- When photon energy increases, Max k.e. increases.
    (ONLY FOR PARTICULATE THEORY)
    And vice versa.
     
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  11. GarryTheGhost

    GarryTheGhost

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    ohh achaaaa it makes sense now. thx
    but what would happen to the photoelectric current tho
     
  12. anastasia grey113

    anastasia grey113

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    learn the formula I=nhf where I is the current, n is the no. of photons and f is the frequency.
    You can also write it as I=nh3x108/λ so that means
    I is inversely proportional to the wavelength.
    So if n (intensity or no.) is constant and wavelength is reduced, the current will increase.
     
  13. GarryTheGhost

    GarryTheGhost

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    i understand how to do part one, its basically alpha=(Z2-Z1)^2/(Z2+Z1)^2 simple acoustic calculation
    but part 2 I done understand how they did it,
    i'm stuck what do to next?
    this is what i did
    0.012I=Ie^alphax
    alpha is 48 from the table
    but why did the marksheme say that alpha x is (48 *2x) and then they multiplied 0.018 with alpha when Ln (ing) bothsides ik that the 0.018 is the reflection coefficient at the boundary between fat and muscle.
    but the part 2 only asks about the thickness of fat so surely i can just say the that alpha is 48 since its the absorbtion coefficient for fat???
     

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  14. anastasia grey113

    anastasia grey113

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    It's because the ray passes twice through the layer of fat before it's intensity is detected.
    So the expression is more like:
    0.018 x (e^-48 x x*10-2)2
    If you solve it this way you will get the same answer
     
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  15. GarryTheGhost

    GarryTheGhost

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    Can you please explain why are you multiplying 0.018 by e^(-48*10-2)2 i
    I understand e^ part but why is 0.018 being multiplied to it.
    shouldn't it just be this:
    0.012=e^((-48*10-2)2)
    i just dont understand why you're using the coefficient alpha thats been calculated in the first part of the question
    Because itn't the formula:
    I=Io e^-αx
    the Intensity cancel out, but what i dont understand is that why are we using the α and the µ at the same time in this question.
    i'm so confused.:cry:
     
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  16. anastasia grey113

    anastasia grey113

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    okay imagine the beam entering
    There will be three phases.
    1- It will move through the layer of fat. As it moves, it will attenuate and intensity will decrease. (e^48 x*10-2)
    2- It will be reflected at the tissue boundary. But some of it will be transmitted thru the muscle tissue. (0.018 of the total is reflected).
    3- It will move across the layer of fat and will be equally attenuated once more (e^48 x 2*102)
    So you just have to multiply all these three phases in order to find the intensity detected.
    The detected intensity is already given (0.012) so put it's value in the equation which is like.
    Intensity detected = not attenuated x reflected x not attenuated

    In short, some of the beam is also 'transmitted and NOT REFLECTED' at the tissue boundary. In this case only 0.018 of it is reflected. The rest is not so obvio it is not detected.
    To get an accurate value, this should also be considered.
     
  17. Hamnah Zahoor

    Hamnah Zahoor

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    Can anyone help me in the following questions from oct/nov/2017/13 Q36 Q37 Untitled.png

    Untitled.png11.png
     
  18. MShaheerUddin

    MShaheerUddin

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  19. kity way

    kity way

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    can someone plz help in b) the ans is 0.85 and 0.51
     

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  20. dumbledore.

    dumbledore.

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    One possible nuclear fission reaction is
    235 1 U + 092 n 141 1
    56 0 Ba + 92
    36 Kr + 3 n + energy.
    Barium-141 (141
    56 Ba) and krypton-92 (92
    36 Kr) are both β-emitters.
    Barium-141 has a half-life of 18 minutes and a decay constant of 6.4 × 10–4 s–1.
    The half-life of krypton-92 is 3.0 seconds.
    (b) A mass of 1.2 g of uranium-235 undergoes this nuclear reaction in a very short time
    (a few nanoseconds).
    (i) Calculate the number of barium-141 nuclei that are present immediately after the
    reaction has been completed.
    Ans is 3.1×10^21
    (ii) Using your answer in (b)(i), calculate the total activity of the barium-141 and the
    krypton-92 a time of 1.0 hours after the fission reaction has taken place.
    activity
    Ms says krypton activity is negligible . Why?
    0c13/43
     
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  21. Hamnah Zahoor

    Hamnah Zahoor

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    I consider when the amplitude increases so does the resultant amplitude/displacement thus more or greater constructive interference occurs resulting in more brighter fringes produced. In other case A^2 is directly proportional to Intensity thus increase in A will ultimately inc. the Intensity of the bright fringes no effect on any other results.
     
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