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German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) to the uncertainty in its velocity ( Δ v ) Δ x ≥ h 4 π m Δ v where h is Planck's constant and m is the mass of the object. The mass of an electron is 9.11 × 10 − 31 kg. What is the uncertainty in the position of an electron moving at 2.00 × 10 6 m/s with an uncertainty of Δ v = 0.01 × 10 6 m/s ?

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According to the information given, the Heisenberg uncertainty principle would be given by the relationship

[tex]\Delta x \Delta v \geq \frac{h}{4\pi m}[/tex]

Here,

h = Planck's constant

[tex]\Delta v[/tex] = Uncertainty in velocity of object

[tex]\Delta x[/tex] = Uncertainty in position of object

m = Mass of object

Rearranging to find the position

[tex]\Delta x \geq \frac{h}{4\pi m\Delta v}[/tex]

Replacing with our values we have,

[tex]\Delta x \geq \frac{6.625*10^{-34}m^2\cdot kg/s}{4\pi (9.1*10^{-31}kg)(0.01*10^6m/s)}[/tex]

[tex]\Delta x \geq 5.79*10^{-9}m[/tex]

Therefore the uncertainty in position of electron is [tex]5.79*10^{-9}m[/tex]

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