Heisenbergs uncertainity principle for an electron

[DODGEVIPER13]

Homework Statement

The speed of an electron is measured to within an uncertainty of 2e4 m/s. What is the size of the smallest region of space in which the electron can be confined?

Homework Equations

Diracs Constant=ΔxΔp
p=mv

The Attempt at a Solution

what I did was (Diracs Constant)/mv = Δx, (1.054560653e-34)/((9.109e-31)(2e4))=6nm this is this incorrect but I don't understand why?

 

[Rajini]

Homework Statement

The speed of an electron is measured to within an uncertainty of 2e4 m/s. What is the size of the smallest region of space in which the electron can be confined?

Homework Equations

Diracs Constant=ΔxΔp
p=mv

The Attempt at a Solution

what I did was (Diracs Constant)/mv = Δx, (1.054560653e-34)/((9.109e-31)(2e4))=6nm this is this incorrect but I don't understand why?

Hi,
I am not sure, I may be wrong.
Use Heisenberg uncertainty principle:
$$\Delta x \Delta p ≥ \hbar/2.$$
The given value is not the speed of electron. It is the uncertainty of speed.
Got it?

 

[sweet springs]

Hi. Not value but only order of value has meaning in this situation. Order of nano meter seems fine. What is the 'correct' answer you have got?

 

[DODGEVIPER13]

Sweet springs the correct answer is 5.8nm what I get is 6nm I know it's so close but it's not what I'm getting. Rajini given the uncertainty of speed how then would I find speed go the electron so I can find p.

 

[DODGEVIPER13]

Ok I'm starting to think I got the right answer after using wolfram I get 5.78 I am starting to think my calculator has some kind of rounding algorithm