# Percentage of Uncertainty

1. Dec 10, 2004

### 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?

2. Dec 10, 2004

### Tide

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.

3. Dec 10, 2004

### CollectiveRocker

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.

4. Dec 10, 2004

### CollectiveRocker

Any ideas guys?

5. Dec 10, 2004

### NateTG

Can you give some more context with this question? Specifically, is it about Quantum Mechanics and Heisenberg's Uncertainty Principle, or something else?

6. Dec 10, 2004

### CollectiveRocker

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?

7. Dec 10, 2004

### NateTG

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.

8. Dec 10, 2004

### CollectiveRocker

Doesn't that mean that my percentage in my uncertainty is 100%?

9. Dec 10, 2004

### dextercioby

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.

10. Dec 10, 2004

### CollectiveRocker

the uncertainty of position is .100 nm

11. Dec 10, 2004

### dextercioby

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.

12. Dec 10, 2004

### CollectiveRocker

The formula for KE = (p^2)/2m
k = 2pi/lambda, and delta p = h/lambda

13. Dec 10, 2004

### NateTG

I also get less than 1% uncertainty.

14. Dec 10, 2004

### CollectiveRocker

with what formula

15. Dec 10, 2004

### dextercioby

Nope,delta p_x is given by the Heisenberg (not de Broglie) formula wrt to h and delta x

16. Dec 10, 2004

### CollectiveRocker

for delta p do you get 6.626 * 10^-24?

17. Dec 10, 2004

### dextercioby

Yes.This thread is getting annoyingly long. :tongue2:

18. Dec 10, 2004

### CollectiveRocker

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

19. Dec 10, 2004

### dextercioby

Pay attention.It's not difficult at all:
$$E=\frac{p^{2}}{2m}$$ From which $$p=\sqrt{2mE}$$.Plug in the numbers,and find the result.If u don't know,learn that
$$m=9.1\cdot 10^{-31} kg$$.

Express the enrgy in Joules,compute the momentum and from there,devide the number u got for $\Delta p$ through the number u got for "p" and express the final result in terms of procents.

Daniel.

20. Dec 10, 2004