Answer :
Final answer:
The uncertainty in the position of an electron moving at 6.00×10⁶ m/s with a 1% uncertainty in speed is calculated using the Heisenberg uncertainty principle, yielding a result of approximately 9.65×10⁻¹ meters.
Explanation:
To calculate the uncertainty in the position, Δx, of an electron using the given speed and its uncertainty, we apply the Heisenberg uncertainty principle, which states that the product of the uncertainties in position and momentum of a particle can never be smaller than a certain value expressed by ℏ/2, where ℏ is the reduced Planck's constant.
The momentum, p, of an electron is given by its mass, m, times its velocity, v, hence the uncertainty in momentum, Δp, can be calculated as the product of the mass of the electron and the uncertainty in velocity, Δv.
First, calculate the uncertainty in momentum:
- Δp = m * Δv
- Δp = (9.109×10⁻³ⁱ kg) * (6.00×10⁴ m/s)
- Δp = 5.465×10⁻² kg⋅m/s
Then, use the Heisenberg uncertainty principle to find the uncertainty in position:
- Δx ≥ ℏ/(2 * Δp)
- Δx ≥ (1.054571800×10⁻4⁴ m²kg/s) / (2 * 5.465×10⁻²3 kg⋅m/s)
- Δx ≥ 9.65×10⁻¹ m
Therefore, the uncertainty in the position of an electron moving near the nucleus at 6.00×10⁶ m/s, with a relative uncertainty in speed of 1%, is approximately 9.65×10⁻¹ meters.
