# Percentage of Uncertainty

by CollectiveRocker
Tags: percentage, uncertainty
 P: 137 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?
 Sci Advisor HW Helper P: 3,144 To find the percentage change divide the change in momentum by the (starting) momentum and multiply by 100%. The change in momentum has units of momentum but the percentage change has no units so they are not the same.
 P: 137 The thing which I've discovered is that when I use that formula, I only end up with 100% back again. Are delta p and p supposed to be the same? For I've taken the things I know, the uncertainty in position and charge, and rearranged them in order to solve for things which I don't know.
 P: 137 Percentage of Uncertainty Any ideas guys?
HW Helper
P: 2,537
 Quote by CollectiveRocker 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?
 P: 137 It is about Heisenberg's Uncertainty Principle. We are given the uncertainty for the position of the 1 KeV electron, and we're asked to find the percentage of uncertainty in it's momentum. Now, I've already found the uncertainty of momentum. However, when I use that answer and solve for momentum, both mometum and the uncertainty in momentum are equal. Thus when I use the % formula: (delta p *100%)/p, I end up with 100% as my percentage. What am I doing wrong?
HW Helper
P: 2,537
 Quote by CollectiveRocker Thus when I use the % formula: (delta p *100%)/p, I end up with 100% as my percentage. What am I doing wrong?
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.
 P: 137 Doesn't that mean that my percentage in my uncertainty is 100%?
HW Helper
P: 11,928
 Quote by CollectiveRocker Doesn't that mean that my percentage in my uncertainty is 100%?
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.

Daniel.
 P: 137 the uncertainty of position is .100 nm
HW Helper
P: 11,928
 Quote by CollectiveRocker the uncertainty of position is .100 nm
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%.

Daniel.
 P: 137 The formula for KE = (p^2)/2m k = 2pi/lambda, and delta p = h/lambda
 Sci Advisor HW Helper P: 2,537 I also get less than 1% uncertainty.
 P: 137 with what formula
HW Helper
P: 11,928
 Quote by CollectiveRocker with what formula
Nope,delta p_x is given by the Heisenberg (not de Broglie) formula wrt to h and delta x
 P: 137 for delta p do you get 6.626 * 10^-24?