Determining initial velocity of an electron

[ZedCar]

Homework Statement

Consider an electron that has its position measured to within a nuclear radius in a laboratory. What is the uncertainty in its position uncertainty one day later?

Homework Equations

The Attempt at a Solution

What equation should I use to determine the initial velocity of the electron?

I believe this would be the first thing I should determine in order to answer the question.

Thanks!

 

[voko]

You need to use the uncertainty principle.

 

[ZedCar]

You need to use the uncertainty principle.

I used

ΔxΔ(mv) = h/2Pi

(10^-14) (9.1 x 10^-31) (Δv) = h/2Pi

Δv = 1.16 x 10^10 m/s

Am I going about this the right way?

What way do I go now about solving the original question?

Do I simply multiply the answer for Δv by (3600x24) i.e. the number of seconds in a day?

If I do this I get an answer of Δx = 1.00 x 10^15 m

Thanks

 

[ZedCar]

Should I instead be using the relativistic equation to find the uncertainty in the momentum, and not p=mv as I done above?

 

[voko]

Yes, you obviously should take relativistic effects into account. Other than that, I think you are on the right track.

 

[ZedCar]

Here is a similar question:
Specifically the *2nd* example
http://www.people.vcu.edu/~rgowdy/mod/122/xmp4.htm

The solution does not appear to have used the relativistic equation.

Should my solution simply be this but substituting their time for a year (3.15x10^7) with my time for one day?

 

[voko]

Well, there are two ways of looking at that. One is that since the velocity thus obtained is greater than the speed of light, the electron simply cannot be localized to within a nuclear radius. Another is that relativity applies and the resultant velocity is less than c anyway. In the latter case, however, the electron's energy is probably going to be greater than many electron rest masses, which means electrons can be created out of nowhere.