# How do I calculate uncertainties for this experiment?

1. Nov 20, 2016

### henry wang

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.
2. Relevant equations

The plotted equation is $$eV=h\frac{c}{2dsin(\theta)}$$ ,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: $$\Delta x=\frac{c*cos(\theta) \Delta \theta}{2dsin^{2}(\theta)}$$
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.
$$\Delta h=\frac{2eVdcos(\theta) \Delta \theta}{c}$$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.

Last edited: Nov 20, 2016
2. Nov 20, 2016

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}$. 3. Nov 20, 2016 ### mfb ### Staff: Mentor 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. 4. Nov 20, 2016 ### henry wang 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. 5. Nov 20, 2016 ### henry wang 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. $$\Delta x=\frac{\delta x}{\delta \theta}\Delta \theta$$ Btw, I updated the OP please check it out. 6. Nov 20, 2016 ### mfb ### Staff: Mentor In the way you did the error propagation, you have to use$\Delta \theta## in radian, not in degrees.

7. Nov 20, 2016

### henry wang

Thank you so much! I changed it to radians and the error is much smaller.

8. Nov 20, 2016

### haruspex

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.

9. Nov 20, 2016

Sure thing!