Heisenberg's uncertainty principle question

  • Thread starter karkas
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  • #1
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Homework Statement


The position and momentum of a 1.00 KeV electron are simultaneously determined. If the position is located to within 0.100 nm, what is the percentage of uncertainty in its momentum?

(Arthur Beiser - Concepts of modern Physics, 3rd Part exercise 33)


Homework Equations


[tex]Δx\cdotΔp≥\frac{\hbar}{2}[/tex]


The Attempt at a Solution



[tex]Δx\cdotΔp≥\frac{\hbar}{2}\rightarrow \; Δp≥ \frac{\hbar}{2\cdotΔx}[/tex]
Given that Δx=0,100 nm, we can evaluate the uncertainty in its momentum Δp.
The momentum of the particle should be of roughly equal magnitude, and evaluating the fraction [tex]\frac{Δp}{p}%[/tex] gives me 98%, where the answer is 3,1 %. What did I do wrong?
 

Answers and Replies

  • #2
MathematicalPhysicist
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What did you plug for p?

It should be p=1[Kev] /c where c is the speed of light.
 
  • #3
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Yep.

[tex]K=\sqrt{m_{0}^{2}c^{4}+p^{2}c^{2}}≈pc[/tex]
so
[tex]p=\frac{1,6\cdot10^{-16} J}{3\cdot10^{8} \frac{m}{s}}=0,533 \cdot 10^{-24} kg\cdot\frac{m}{s}[/tex]
and because [tex]Δp=\frac{1,05\cdot10^{-34}J\cdot s}{2\cdot 10^{-10}m}=0,525\cdot 10^{-24} kg\cdot \frac{m}{s} [/tex]
we get
[tex]\frac{Δp}{p}≈0,98[/tex]

:(
 
  • #4
132
1
Could I get a bumpity-bump 'cause these exercises are crucial for the finals?
 

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