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How do you go about finding the percentage in the uncertainty of momentum, if you already know the momentum and delta momentum? I'm asking because both of them end up being the same; thus I'm wondering if I'm terribly wrong. Any advice?
Can you give some more context with this question? Specifically, is it about Quantum Mechanics and Heisenberg's Uncertainty Principle, or something else?CollectiveRocker said:How do you go about finding the percentage in the uncertainty of momentum, if you already know the momentum and delta momentum? I'm asking because both of them end up being the same; thus I'm wondering if I'm terribly wrong. Any advice?
This is ceratinly not my area of expertise, but why do you think having the uncertainty equal to the momentum is an incorrect answer? From what I understand the uncertainty can be larger than the momentum as well.CollectiveRocker said:Thus when I use the % formula: (delta p *100%)/p, I end up with 100% as my percentage. What am I doing wrong?
Why not??As far as the calculations u made are correct,then that should be it.But i'd like to see all the numbers,though.U say the KE of the electron is 1KeV.Please give us the uncertainty in distance.CollectiveRocker said:Doesn't that mean that my percentage in my uncertainty is 100%?
I'm sorry to say,that,but you screwed up the numbers.Did u use the correct (nonrelativistic) formula for the momentum in terms of the KE??If so,combined with Heisenberg formula u should be getting less than 1%.CollectiveRocker said:the uncertainty of position is .100 nm
Nope,delta p_x is given by the Heisenberg (not de Broglie) formula wrt to h and delta xCollectiveRocker said:with what formula
Yes.This thread is getting annoyingly long. :tongue2:CollectiveRocker said:for delta p do you get 6.626 * 10^-24?
Pay attention.It's not difficult at all:CollectiveRocker said:I'm sorry to keep on testing your patience. So then we solve for k using k = 2pi/lambda, because lambda = delta x, and I get 6.283 * 10^10