- #1

- 555

- 19

i'm solving some quite simple problems using the uncertainty principle, but I don't have access to the solutions and I really don't have a feel for what a 'sensible' answer is... When finding the minimum uncertainty in velocity, I end up with things greater than the speed of light, so I want to check i'm not making an error somewhere.

## Homework Statement

Suppose that an electron is trapped in a small region and the uncertainty in it's position is [itex]10^{-15}[/itex]m. What is the minimum uncertainty in the particles momentum? What uncertainty in the electrons velocity does this correspond to?

## Homework Equations

## The Attempt at a Solution

[itex](\Delta y)(\Delta p_{y}) \geq \frac{h}{4 \pi}[/itex]

Since we're dealing the minimum uncertainty, we can equate the two.

[itex](\Delta p_{y}) = \frac{h}{4 \pi (\Delta y)}[/itex] = 5.27 x 10

^{-20}kgms

^{-1}

[itex]p = mv[/itex]

[itex]\Delta p_{y} = m (\Delta v_{y})[/itex]

[itex]\Delta v_{y} = \frac{\Delta p_{y}}{m}[/itex] = 5.79 x 10

^{10}ms

^{-1}

This answer just seems absurd, maybe it's a consequence of the accuracy we know the electrons position to, but like I said I have no way of checking what a sensible answer is.

So, have I gone wrong somewhere?

Thanks for any help you can give.