Question

Find the uncertainty in position (in nm) associated with an electron that is moving with a velocity of 572 km/s. The uncertainty in the velocity is 5.00%. The electron rest mass is 9.11 x 10-31 kg.

Answer 1

The uncertainty in position of an electron can be calculated using the Heisenberg Uncertainty Principle. For the given velocity of 572 km/s and an uncertainty in velocity of 5.00%, the uncertainty in position is approximately 1.847 fm.

Step-by-step explanation:

The uncertainty in position (Δx) of an electron is related to the uncertainty in velocity (Δv) by the Heisenberg Uncertainty Principle. The principle states that the product of Δx and Δv must be greater than or equal to the reduced Planck's constant (ħ/2). Mathematically, Δx Δv ≥ ħ/2.

Given that the uncertainty in velocity is 5.00% of the actual velocity, we can calculate Δv = 0.0500 × 572 km/s = 28.6 km/s. Converting this to meters per second, we get Δv = 28.6 km/s × (1000 m/km) × (1 s/1000 ms) = 28,600 m/s.

Now we can use the uncertainty principle equation to calculate the uncertainty in position:

Δx = (ħ/2) / Δv = (1.055 × 10^(-34) kg m²/s) / (2 × 28,600 m/s) = 1.847 × 10^(-39) m or 1.847 fm (femtometers).