Question

German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) (Δx) to the uncertainty in its velocity ( Δ v ) (Δv) Δ x ≥ h 4 π m Δ v Δx≥h4πmΔv where h h is Planck's constant and m m is the mass of the object. The mass of an electron is 9.11 × 10 − 31 kg. 9.11×10−31 kg. What is the uncertainty in the position of an electron moving at 6.00 × 10 6 m/s 6.00×106 m/s with an uncertainty of Δ v = 0.01 × 10 6 m/s ?

Answer 1

Answer:

$5.788\times 10^{-9} m$ is the uncertainty in the position of a moving electron.

Explanation:

Heisenberg's uncertainty principle is given by the equation:

$\Delta x\times m\times \Delta v=\frac{h}{4\pi}$

The mass of an electron = m

Uncertainty in velocity =  Δv

Uncertainty in position =  Δx

h = Planck's constant

We are given:

The mass of an electron = m = $9.11\times 10^{-31} kg$

Uncertainty in velocity =  Δv = $0.01 \times 10^6 m/s$

Uncertainty in position =  Δx

$\Delta x=\frac{h}{4\pi \times m\times \Delta v}$

$=\frac{6.626\times 10^{-34} Js}{4\times 3.14\times 9.11\times 10^{-31} kg\times 0.01 \times 10^6 m/s}$

$=5.788\times 10^{-9} m$

$5.788\times 10^{-9} m$ is the uncertainty in the position of a moving electron.