# Homework Help: Localizing an electron problem (need help!)

1. Dec 11, 2011

### jakepeck

1. The problem statement, all variables and given/known data

Suppose I use visible (say yellow) light to observe an electron.
Roughly speaking, I will then be able to localize the electron to a region
about the size of the wavelength of the light. According to the
uncertainty principle, I am now limited as to how well I know the momentum
of the electron. If the maximum momentum that the electron can have is
roughly equal to the uncertainty of the momentum, about how long will I
have to wait so that the electron could be anyplace in the City of Ann
Arbor? To do this, you will need to estimate the size of Ann Arbor, and
calculate how long it would take the electron to go a distance about that
size.

2. Relevant equations

I chose D = Rt and A = pi(R^2)

3. The attempt at a solution

I just decided to use an area equation, pretending Ann Arbor is a circular town, using a given area of 27 miles.

Then I just used D = rt to find the time it would take the electron to go that distance, not completely sure if that was the right thing to do. I used c as the speed assuming the electron goes that speed.... Is that maybe where my problem is wrong?

my answer was t = 3.15 x 10^(-5) seconds.

2. Dec 11, 2011

### e.bar.goum

That's certainly it! Electrons can't travel at c! Use the uncertainty principle to find the uncertainty in the momentum, and thus the uncertainty in the electrons speed, and go from there.

3. Dec 11, 2011

### jakepeck

Am I supposed to use deltaXdeltaP = h ? What equation am I supposed to use? I see a lot of different equations for this.

4. Dec 11, 2011

### e.bar.goum

Nearly,

Delta x Delta p >= hbar/2

5. Dec 11, 2011

### jakepeck

Alright, I used deltaXdeltaP >= hbar/2 and brought it down to

9436.4 meters(9.109x10^(-31)kg)v>=hbar/2

which gave me v >= 6.134 x 10^(-9) m/s

so i plugged that back into D = rt

9436.4 = (6.134x10^(-9))t

t = 1.54 x 10^12 seconds

Is this right?

6. Dec 11, 2011

### e.bar.goum

9436.4 meters doesn't sound like the wavelength of yellow light to me!

7. Dec 11, 2011

### jakepeck

I used the distance of ann arbor... But I guess im supposed to use the wavelength of the light in this case? 580 nm for yellow light?

8. Dec 11, 2011

### e.bar.goum

Well, the uncertainty in the position is the wavelength of the light. So, yes.

Edited to add - this is a really strange question, I have to say.

9. Dec 11, 2011

### jakepeck

ok this time I did

5.8 x 10^(-4)(9.109 x 10^(-31))v >= hbar

which lead me to v >= 9.975 x 10^(-24)

plugged into D = rt

9436.4 m = (9.975 x 10^(-24))t

t= 9.46 x10^(26) s

And yes, this is my professor at University of Michigan, a weird dude.

10. Dec 11, 2011

### e.bar.goum

5.8 x 10^(-4) is not equal to 580nm

11. Dec 11, 2011

### jakepeck

I thought there was 1,000,000 nm in a meter? 580 nm X 1 meter / 1,000,000 nm = 5.8 x 10^(-4) m

12. Dec 11, 2011

### e.bar.goum

Try another 3 orders of magnitude. 1nm = 1*10^-9 m

13. Dec 11, 2011

### jakepeck

ahahha oops okay....

now after plugging everything in the same way but with the new wavelength....

v >= 99.79 m/s

which leads me to t = 94.56 seconds

Did I finally make it?

14. Dec 11, 2011

### e.bar.goum

No idea. I haven't calculated it myself!

Does the answer seem reasonable to you?

15. Dec 11, 2011

### jakepeck

I think it sounds pretty reasonable, as Ann Arbor is decently sized. 94 seconds sounds about right.