Heisenbergs uncertainity principle for an electron

  • #1
DODGEVIPER13
672
0

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 dont understand why?
 

Answers and Replies

  • #2
Rajini
619
2

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 dont understand why?
Hi,
I am not sure, I may be wrong.
Use Heisenberg uncertainty principle:
[tex]\Delta x \Delta p ≥ \hbar/2.
[/tex]
The given value is not the speed of electron. It is the uncertainty of speed.
Got it?
 
  • #3
sweet springs
1,225
75
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?
 
  • #4
DODGEVIPER13
672
0
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.
 
  • #5
DODGEVIPER13
672
0
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
 

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