Heisenberg's Uncertainty Principle

[BOAS]

Hello,

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 10^{-15}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

(\Delta y)(\Delta p_{y}) \geq \frac{h}{4 \pi}

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

(\Delta p_{y}) = \frac{h}{4 \pi (\Delta y)} = 5.27 x 10^-20 kgms^-1

p = mv

\Delta p_{y} = m (\Delta v_{y})

\Delta v_{y} = \frac{\Delta p_{y}}{m} = 5.79 x 10¹⁰ 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.

 

[dauto]

Your mistake is to use the formula p = mv which is the non-relativistic formula. Try using p = γmv, where γ = (1 - v²/c²)^-1/2.

 

[BOAS]

Your mistake is to use the formula p = mv which is the non-relativistic formula. Try using p = γmv, where γ = (1 - v²/c²)^-1/2.

Aha! Thanks, I was looking at an old example from my notes but evidently it was intended to show why the Heisenberg uncertainty principle is negligible at the macroscopic level of a ping pong ball.