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How do I calculate uncertainties for this experiment?

  1. Nov 20, 2016 #1
    1. The problem statement, all variables and given/known data
    I used error propagation equation to calculate errors in x and y for each individual data. Error in y is negligible. My goal is to find the uncertainty in the gradient. Capture.PNG
    2. Relevant equations

    The plotted equation is [tex]eV=h\frac{c}{2dsin(\theta)}[/tex] ,where Plank's constant, h, is the gradient, the independent variables eV and the dependent variable is theta. Also x=c/(2dsin(theta) and y=eV.
    The dominant error is in the theta, and thus error in x is: [tex]\Delta x=\frac{c*cos(\theta) \Delta \theta}{2dsin^{2}(\theta)}[/tex]
    c=3*10^8m, cos(theta)=1, d is about 10^-10m, sin(theta) is about 0.1 and dtheta=0.2 degrees.
    This yields an extremely big error in x.

    3. The attempt at a solution
    My original thought was to fit lines of max and min permitted gradient using errors in x and y. However, since the error in x is so large, and the x and y-axis are in log scale, I cannot manually fit lines.
    Update: Therefore I rearranged the above equation to isolate h, calculated the uncertainties in h for each data points and took its average.
    [tex]\Delta h=\frac{2eVdcos(\theta) \Delta \theta}{c}[/tex]The planks constant, h, was found to be 6.7*10-34Js, and the average uncertainty in h was found to be +-1.51^-33Js.
    Is this a reasonable approach?

    Update 2: After corrected dtheta from degrees to radians.
    Caspture.PNG
     
    Last edited: Nov 20, 2016
  2. jcsd
  3. Nov 20, 2016 #2
    Are you sure that your units are correct? ##1.0e-15## seems to be very small energies. I have never heard about any experiment operating at such small energies. Also, it seems like your expression for ##\Delta x## doesn't have the correct unit. I assume that ##\Delta x# should be a distance but your expression gives ##s^{-1}##.
     
  4. Nov 20, 2016 #3

    mfb

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    Just by looking at the data: either your uncertainties are all extremely correlated (and I don't see why they should be) or you overestimate your uncertainties massively.

    I would expect the second case (especially as your wavelength is certainly not negative), but without a description of your experiment it is hard to figure out what went wrong.
     
  5. Nov 20, 2016 #4
    Sorry I should've specified x is just a dummy variable, it is c/wavelength with unit 1/s. The energy in this experiment is x-ray photon energy.
     
  6. Nov 20, 2016 #5
    The reason why error in x is so big is because inter-crystal plane distance, d, is very small and c is very large. The uncertainty in x was calculated using the error propagation equation. [tex]\Delta x=\frac{\delta x}{\delta \theta}\Delta \theta[/tex]
    Btw, I updated the OP please check it out.
     
  7. Nov 20, 2016 #6

    mfb

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    In the way you did the error propagation, you have to use ##\Delta \theta## in radian, not in degrees.
     
  8. Nov 20, 2016 #7
    Thank you so much! I changed it to radians and the error is much smaller.
     
  9. Nov 20, 2016 #8

    haruspex

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    Would you post the new graph, please? It is hard to judge what it will look like with the horizontal scale blown up by a factor of about 3000.
     
  10. Nov 20, 2016 #9
    Sure thing!
     
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