# Energy resolution of triple-axis spectrometer

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1. Dec 30, 2013

### revbrapok

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.

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. Dec 30, 2013

### 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.

3. Dec 31, 2013

### M Quack

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. Dec 31, 2013

### 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.

5. Dec 31, 2013

### M Quack

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. Dec 31, 2013

### Staff: Mentor

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