The mass of an electron is 9.11× 10–31 kg. What is the uncertainty in the position of an electron moving at 5.00 × 106 m/s with an uncertainty of Δv = 0.01 × 106 m/s?
Δx ≥ h/(4π * Δp)
deltap=mass*deltaV; h is planck's constant.
To find the uncertainty in the position of an electron, we can use the Heisenberg uncertainty principle, which states: Δx * Δp ≥ h/4π, where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant (6.626 × 10^-34 J s).
First, let's find the uncertainty in momentum (Δp):
Δp = m * Δv
where m is the mass of the electron (9.11× 10–31 kg) and Δv is the uncertainty in velocity (0.01 × 10^6 m/s).
Plugging in the values:
Δp = (9.11× 10–31 kg) * (0.01 × 10^6 m/s)
= 9.11 × 10^ -25 kg m/s
Next, we can calculate the uncertainty in position (Δx):
Δx * Δp = h/4π
Rearranging the equation:
Δx = h / (4π * Δp)
Plugging in the values:
Δx = (6.626 × 10^-34 J s) / (4π * 9.11 × 10^-25 kg m/s)
Calculating Δx:
Δx ≈ 5.70 × 10^-10 m
Therefore, the uncertainty in the position of the electron is approximately 5.70 × 10^-10 meters.