How do I calculate uncertainties for this experiment?

[henry wang]

Homework Statement

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. Homework 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.

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.

 

[eys_physics]

Homework Statement

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. View attachment 109168
2. Homework 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.

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.
Therefore I calculated the uncertainties in h for each data points and took its average. The planks constant, h, was found to be 6.7*10-34Js, and the average uncertainty in h was found to be 1.51^-33.
Is this a reasonable approach?

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}##.

 

[mfb]

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.

 

[henry wang]

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}##.

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.

 

[henry wang]

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.

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.

 

[mfb]

In the way you did the error propagation, you have to use $\Delta \theta$ in radian, not in degrees.

 

[henry wang]

In the way you did the error propagation, you have to use $\Delta \theta$ in radian, not in degrees.

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

 

[haruspex]

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

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.

 

[henry wang]

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.

Sure thing!