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

`\Delta x\geq 5.78793* 10^(-9)\ m`

Step-by-step explanation:

h = Planck's constant =
`6.626* 10^(-34)\ m^2kg/s`

m = Mass of electron =
`9.11* 10^(-31)\ kg`

`\Delta v` = Speed of electron =
`0.01* 10^6\ m/s`

According to the Heisenberg uncertainty principle

`m\Delta x\Delta V\geq (h)/(4\pi)\\\Rightarrow \Delta x\geq (h)/(4\pi* m* \Delta v)\\\Rightarrow \Delta x\geq (6.626* 10^(-34))/(4\pi* 9.11* 10^(-31)* 0.01* 10^6)\\\Rightarrow \Delta x\geq 5.78793* 10^(-9)\ m`

Therefore,
`\Delta x\geq 5.78793* 10^(-9)\ m`