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Percentage of Uncertainty

by CollectiveRocker
Tags: percentage, uncertainty
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CollectiveRocker
#1
Dec10-04, 12:14 AM
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?
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#2
Dec10-04, 06:37 AM
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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.
CollectiveRocker
#3
Dec10-04, 08:42 AM
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.

CollectiveRocker
#4
Dec10-04, 12:30 PM
P: 137
Percentage of Uncertainty

Any ideas guys?
NateTG
#5
Dec10-04, 12:31 PM
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Quote 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?
CollectiveRocker
#6
Dec10-04, 02:13 PM
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?
NateTG
#7
Dec10-04, 02:39 PM
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Quote 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.
CollectiveRocker
#8
Dec10-04, 02:41 PM
P: 137
Doesn't that mean that my percentage in my uncertainty is 100%?
dextercioby
#9
Dec10-04, 03:07 PM
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Quote 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.
CollectiveRocker
#10
Dec10-04, 03:10 PM
P: 137
the uncertainty of position is .100 nm
dextercioby
#11
Dec10-04, 03:28 PM
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Quote 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.
CollectiveRocker
#12
Dec10-04, 03:36 PM
P: 137
The formula for KE = (p^2)/2m
k = 2pi/lambda, and delta p = h/lambda
NateTG
#13
Dec10-04, 03:45 PM
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I also get less than 1% uncertainty.
CollectiveRocker
#14
Dec10-04, 03:53 PM
P: 137
with what formula
dextercioby
#15
Dec10-04, 04:00 PM
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Quote 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
CollectiveRocker
#16
Dec10-04, 04:03 PM
P: 137
for delta p do you get 6.626 * 10^-24?
dextercioby
#17
Dec10-04, 04:06 PM
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Quote Quote by CollectiveRocker
for delta p do you get 6.626 * 10^-24?
Yes.This thread is getting annoyingly long.
CollectiveRocker
#18
Dec10-04, 04:22 PM
P: 137
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


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