Question

An electron has an uncertainty in its position of 552pm. What is the uncertainty in its velocity?

Answer 1

To calculate the uncertainty in the electron's velocity with an uncertainty in its position of 552pm, you apply the Heisenberg Uncertainty Principle, using the reduced Planck constant and the electron's mass to find the uncertainty in velocity (Δv).

Step-by-step explanation:

To determine the uncertainty in the velocity of an electron with an uncertainty in its position of 552pm, we utilize the Heisenberg Uncertainty Principle which states that the product of the uncertainties in position (Δx) and momentum (Δp) must be greater than or equal to ħ/2, where ħ is the reduced Planck constant (1.055 × 10-34 kg m2/s).

The uncertainty in position given as 552pm (or 552 × 10-12 m) together with the uncertainty in momentum which is the mass of the electron (me, approximately 9.11 × 10-31 kg) times the uncertainty in velocity (Δv), gives us the following relation:

ħ/2 ≤ Δx × me × Δv

Plugging in the numbers, we solve for Δv:

1.055 × 10-34 kg m2/s / 2 ≤ (552 × 10-12 m) × (9.11 × 10-31 kg) × Δv

From this, the uncertainty in the electron's velocity can be calculated, providing an answer to the student's question.

Answer 2

Make an equation like that :

Δx×Δmu≥h:4π

This will help you to find Dv, you just need to rearrange uncertainty equation.