Energy resolution of tripleaxis spectrometerby revbrapok Tags: energy, energy resolution, resolution, spectrometer, tripleaxis 

#1
Dec3013, 03:27 AM

P: 1

Hi,
I have encountered the problem of energy resolution of neutron tripleaxis spectrometer, which we haven't covered during our solid state physics lectures. 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
Dec3013, 08:43 AM

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P: 10,824

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. 



#3
Dec3113, 03:40 AM

P: 638

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. 



#4
Dec3113, 01:04 PM

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Energy resolution of tripleaxis spectrometer 



#5
Dec3113, 01:30 PM

P: 638

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 :)) 



#6
Dec3113, 02:00 PM

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P: 10,824

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


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