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German physicist Werner Heisenberg related the uncertainty of an object's position (Δx) to the uncertainty in its velocity (Δ????) Δx≥ℎ4????mΔ???? 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 4.00×106 m/s with an uncertainty of Δ????=0.01×106 m/s?

Respuesta :

Answer:

[tex]\Delta x = 5.47 \times 10^{-9} m[/tex]

Explanation:

As we know by the principle of uncertainty that the product of uncertainty in position and uncertainty in momentum is given as

[tex]\Delta x \times \Delta P = \frac{h}{4\pi}[/tex]

so here we know that

[tex]\Delta v = 0.01 \times 10^6 m/s[/tex]

[tex]m = 9.11 \times 10^{-31} kg[/tex]

so we have

[tex]\Delta x \times (9.11 \times 10^{-31})(0.01 \times 10^6) = \frac{6.26 \times 10^{-34}}{4\pi}[/tex]

[tex]\Delta x = 5.47 \times 10^{-9} m[/tex]

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