Heisenbergs uncertainity principle for an electron

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SUMMARY

The discussion centers on applying the Heisenberg Uncertainty Principle to calculate the confinement of an electron based on its speed uncertainty. The uncertainty in speed is given as 2e4 m/s, and the calculation involves Dirac's constant and momentum (p=mv). The initial calculation yielded 6 nm, but the correct value is determined to be approximately 5.78 nm. This discrepancy is attributed to rounding errors in calculations.

PREREQUISITES
  • Understanding of the Heisenberg Uncertainty Principle
  • Familiarity with Dirac's constant (ħ)
  • Basic knowledge of momentum (p=mv)
  • Proficiency in performing calculations involving scientific notation
NEXT STEPS
  • Study the Heisenberg Uncertainty Principle in detail
  • Learn about Dirac's constant and its applications in quantum mechanics
  • Explore advanced calculations involving momentum and uncertainty
  • Investigate rounding errors in scientific calculations and their implications
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Students studying quantum mechanics, physicists interested in particle behavior, and educators teaching the principles of uncertainty in physics.

DODGEVIPER13
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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?
 
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DODGEVIPER13 said:

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.<br />
The given value is not the speed of electron. It is the uncertainty of speed.
Got it?
 
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?
 
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.
 
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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