Question

German physicist Werner Heisenberg related the uncertainty of an object's position (Δx) to the uncertainty in its velocity (Δv) Δx ≥ h/4πmΔv 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 1.00×10⁶ m/s with an uncertainty of Δv=0.01×10⁶ m/s? Δx ≥ ______.

Answer 1

Answer:

$\Delta x\geq 5.78793\times 10^{-9}\ m$

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

$\Delta v$ = Speed of electron = $0.01\times 10^6\ m/s$

According to the Heisenberg uncertainty principle

$m\Delta x\Delta V\geq \frac{h}{4\pi}\\\Rightarrow \Delta x\geq \frac{h}{4\pi\times m\times \Delta v}\\\Rightarrow \Delta x\geq \frac{6.626\times 10^{-34}}{4\pi\times 9.11\times 10^{-31}\times 0.01\times 10^6}\\\Rightarrow \Delta x\geq 5.78793\times 10^{-9}\ m$

Therefore, $\Delta x\geq 5.78793\times 10^{-9}\ m$