Heisenberg uncertainty relation

Question

An electron is located within an interval of 0.247 nm0.247 nm in the north-south direction. What is the minimum uncertainty Δ𝑣Δ⁢v in the electron's velocity in that direction?

The Heisenberg uncertainty relation is given different forms in different textbooks. Use the form employing ≥ℎ4π≥h4⁢π.

Answer 1

The Heisenberg uncertainty relation states that the uncertainty in position (Δx) and the uncertainty in momentum (Δp) are related by the equation:

Δx * Δp ≥ / 4π

The momentum if an electron is given by the equation p=mv, where m is the mass of the electron and v is the velocity. In this case, we are looking for the minimum uncertainty in the velocity of the electron (Δv) in the north-south direction.

To solve for Δv, we can rearrange the Heisenberg uncertainty relation to solve for Δp:

Δp = h /4π * (1/Δx)

We can then use the equation for momentum (p = mv) to solve for Δv:

Δv = Δp / m

Plugging in the values given in the problem, we get:

Δv = h / 4π * (1/0.247) / m

Δv = 3.2 x 10⁸ m/s

Thus, the minimum uncertainty in the electron's velocity in the north-south direction is 3.2 x 10⁸ m/s.