Question

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 1.00 × 10 6 m/s with an uncertainty of Δ v = 0.01 × 10 6 m/s ?

Answer 1

Answer:

The uncertainty in the position of the electron is $5.79x10^{-9}m$

Explanation:

The Heisenberg uncertainty principle is defined as:

$\Lambda p\Lambda x$ ≥ $\frac{h}{4 \pi}$  (1)

Where $\Lambda p$ is the uncertainty in momentum, $\Lambda x$ is the uncertainty in position and h is the Planck's constant.

The momentum is defined as:

$p =mv$  (2)

Therefore, equation 2 can be replaced in equation 1

$\Lambda (mv) \Lambda x$ ≥ $\frac{h}{4 \pi}$

Since, the mass of the electron is constant, v will be the one with an associated uncertainty.

$m \Lambda v \Lambda x$ ≥ $\frac{h}{4 \pi}$ (3)

Then, $\Lambda x$ can be isolated from equation 3

$\Lambda x$ ≥ $\frac{h}{m \Lambda v 4 \pi}$  (4)

$\Lambda x = \frac{6.626x10^{-34}J.s}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) 4 \pi}$

But $1J = Kg.m^{2}/s^{2}$

$\Lambda x = \frac{(6.624x10^{-34} Kg.m^{2}/s^{2}.s)}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) (4 \pi)}$

$\Lambda x = 5.79x10^{-9}m$

Hence, the uncertainty in the position of the electron is  $5.79x10^{-9}m$