Energy resolution of triple-axis spectrometer

  1. Hi,
    I have encountered the problem of energy resolution of neutron triple-axis spectrometer, which we haven't covered during our solid state physics lectures.
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    I don't know where do we get the equation for the energy resolution from and even the numerical calculations in the solution seem odd to me as I am unable to get the same results. Can someone give me an insight into this or better suggest me some literature covering this problem?
     
  2. jcsd
  3. mfb

    Staff: Mentor

    The total resolution comes from the wavelength distribution of the incoming neutrons together with the accurary of the analyzer - that looks fine. The given uncertainties are relative values, so the energy uncertainty gets multiplied with the corresponding energy.
    I would expect a quadratic sum (##\sqrt{a^2+b^2}## instead of a+b) and I don't understand the prefactors of 2, but that could be some (conservative) convention.

    I can confirm the neutron energy, I have no idea where the 5 meV come from.
    The intermediate step in the final calculation looks completely wrong, but the result of 37µeV agrees with the formula.
     
  4. http://en.wikipedia.org/wiki/Propagation_of_uncertainty, look at sections "Simplification" and "Example". With this you should be able to figure out where the "2" comes from.

    The intermediate step should probably read 2 x (13.06 + 8.06) x 10^-3, using the simplification that delta lambda/lambda is pretty much the same for the monochromator and the analyzer.

    I agree with mfb that one should probably use a quadratic sum.

    The 5 meV is just an example of the excitation one might measure with this inelastic neutron scattering experiment.
     
  5. mfb

    Staff: Mentor

    The only "2" I see there is related to a correlation between two uncertain parameters, something we do not have here. And we don't have squared parameters either.
     
  6. E is quadratic in k and thus lambda, and the error propagation is (or should be)

    (delta E)^2 = (delta lambda_0)^2 (d E_0/d lambda_0)^2 + ...

    = (delta lambda_0)^2 (-2 hbar^2 /(2m lambda_0^3)^2 + ...

    = (delta lambda_0/lambda_0 2 E_0)^2 + ...

    correct me if I am wrong. (I have 1/2 bottle of wine as excuse :-))
     
  7. mfb

    Staff: Mentor

    Oh right...
    Okay, that explains the factor of 2.
     
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