How should I calculate the error in my e/m ratio calculation for my homework?

  • Thread starter Thread starter Phystudent91
  • Start date Start date
  • Tags Tags
    Calculation Error
Phystudent91
Messages
6
Reaction score
0

Homework Statement


I'm in the middle of a project finding the e/m ratio of the electron by the use of the formula below. I know all required values and can find a result, I'm unsure how to calculate an error for the calculation.
In the formula below, D, n, A and f are all given quantities (therefore no error/the error can be taken as 0, right?)

Homework Equations



\frac{e}{m} = (32/D2)*(nA2∏)2*(Va/Is2)*(Is/Vs)2

The Attempt at a Solution



The last 2 parts of the equation are where the trouble arises. I have had to measure and record results in two different ways to find values and to plot the 2 graphs - Va/Is2 and Is/Vs - and the software used gives a percentage error in the line.
Do I need to find the error in each measurement for Va and Is etc?

The current way I know to calculate error would mean using the smallest value measured and half of the smallest possible measurement (half the equipment increment). This would result in a stupidly high error, eg:
smallest V measurement was 5V, this is also the smallest increment on the equipment, therefore error is ±2.5V meaning an error in that set of measurements of 50%??
I'm then taught to add the percentage errors... this leads to 350% error.
However, because I know the gradients of the graph, is it acceptable to only use these errors? This would give a total are of ≈10%. Much better!

Which is the correct way of going about this? And if the answer is neither of these, what is the correct way of going about this!?

Thanks in advance!
 
Physics news on Phys.org
Any Help would be appreciated!
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top