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Double slit electron experiment

  1. Mar 17, 2005 #1
    A monochromated beam of neutrons (speed 0.40 m/s) is directed through a double slit with 1.0 mm separation. An array of detectors is placed 10m from the slit.
    a)what is the de Broglie wavelength of the neutrons?
    b)how far off-axis is the first zero-intensity point on the detector array?
    c)explain why we cannot measure which slit a neutron passed through while forming a diffraction pattern.

    for a) i used lambda= h/p and came up with 9.89*10^-7. i hope this is right.

    the rest of the question is a problem. i am lost. the only equations i can find are Dsin(theta) =nlambda (given by my proff) and Dsin(theta)= lambda/2. i know that i can't just use either of these equations on their own, but i am baffled as to what i should use. i'm not looking for anyone to pretty much do the question, i just need a nudge in the right direction and hopefully a simple explanation as to what is actually going on in this situation, and what the question is asking. thanks
     
  2. jcsd
  3. Mar 17, 2005 #2
    for part c think about uncertainty. The only way to detect if a neutron passed through a slit wuld be to use a photon which would end up 'dsitracting' the neutron on its course thus ruining everything
     
  4. Mar 17, 2005 #3
    i think i understand what you are saying for c). the text describes a similar situation, where the detecting particle has uncertainty in its momentum and postion. through calculations they determine that for the dectecting particle:

    (delta P)(delta y) < h/2. this is a clear violation of HUP.

    anyone have any thoughts for b) to help me understand?
     
  5. Mar 17, 2005 #4
    well you have the wavelength of the neutrons so couldn't you simply use the double slit diffraction to get the answer for part b

    [tex] dsin \theta = (m+ \frac{1}{2}) \lambda [/tex]
    and once you find the theta then [tex] tan \theta = \frac{\Delta y}{L} [/tex]
    where d is teh separation slits and ambda is the wavelength and L is the distance to the screen
     
  6. Mar 17, 2005 #5
    i have no clue as to where you came up with dsin(theta)=(m+1/2)(lambda)

    also what does the tan equation mean. i need some help to understand what is going on. what is delta y? i am so lost on this topic. any layman's terms explainations would be so appreciated. thanks.
     
  7. Mar 17, 2005 #6
    have you done double slit diffraction in the past?
    the first equation is the separation of maxima for the double slit patern created for light. Can be used for neutrons too for the idealized case
     
  8. Mar 17, 2005 #7
    isn't dsin(theta)= (m+1/2)lambda for the dark fringes of the double slit. how do i know to use this instead of just dsin(theta)= mlambda?

    also, again, i don't understand where you got the tan equation. it has been a very long time since i did anything with diffraction, and i have one of those proffs who is hardly ever available, and when you do catch him he looks at you like you are stupid for asking a question.

    with the equation you give would m then be one because they ask for the first zero-intensity point?
     
  9. Mar 17, 2005 #8
    part b of your question is
    b)how far off-axis is the first zero-intensity point on the detector array?

    zero intensity means ZERO neutrons/photons/potatoes have landed on th target. (since these are neutrosn it wont be PERFECTLY zero, but i'm assuming that you're not that high a stage yet
    chekc your physics textbook for a more detailed explanation of this topic.

    llook for this expression and understand where the theta is actaully located
     
  10. Mar 17, 2005 #9

    jtbell

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    Staff: Mentor

    Your second equation gives you the locations of the points of maximum intensity. The problem asks you for the location of one of the points of minimum intensity, which are located halfway between the points of maximum intensity. That's where the 1/2 comes from.

    Doesn't your textbook have a diagram of the setup? There should be a long right triangle with the tip at the slits, and the short side at the screen, and the long side along the axis of the setup (running from the slits to the center of the screen, and perpendicular to the screen. The angle at the tip is [itex]\theta[/itex]. The opposite side is [tex]\Delta y[/itex], the distance along the screen from the center to the point in question. The adjacent side is [itex]L[/itex], the distance from the slits to the screen.
     
  11. Dec 6, 2009 #10
    for this same example, how would you find the wave function?
     
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