To determine the uncertainty in the position of an electron given its mass and velocity, we can utilize Heisenberg's Uncertainty Principle. This principle states that the product of the uncertainties in position (Δx) and momentum (Δp) of a particle cannot be smaller than a certain value. Mathematically, it is expressed as:
Δx * Δp ≥ ħ / 2
Here, ħ (h-bar) is the reduced Planck's constant, approximately equal to 1.055 x 10^-34 Js. Let's break down the problem step by step.
Step 1: Calculate the momentum uncertainty (Δp)
The momentum (p) of an electron can be calculated using the formula:
p = m * v
Where:
- m = mass of the electron = 9.1 x 10^-28 g = 9.1 x 10^-31 kg (since 1 g = 10^-3 kg)
- v = velocity of the electron = 3.0 x 10^4 cm/s = 3.0 x 10^2 m/s (since 1 cm = 10^-2 m)
Now, substituting the values:
p = (9.1 x 10^-31 kg) * (3.0 x 10^2 m/s) = 2.73 x 10^-28 kg·m/s
Next, we need to find the uncertainty in momentum (Δp). The uncertainty in velocity (Δv) can be calculated from the given accuracy of the velocity:
Δv = 0.011% of v = 0.011/100 * 3.0 x 10^2 m/s = 3.3 x 10^-2 m/s
Now, we can find Δp using the formula:
Δp = m * Δv
Substituting the values:
Δp = (9.1 x 10^-31 kg) * (3.3 x 10^-2 m/s) = 3.003 x 10^-32 kg·m/s
Step 2: Calculate the uncertainty in position (Δx)
Now that we have Δp, we can use Heisenberg's Uncertainty Principle to find Δx:
Δx ≥ ħ / (2 * Δp)
Substituting the known values:
Δx ≥ (1.055 x 10^-34 Js) / (2 * 3.003 x 10^-32 kg·m/s)
Calculating this gives:
Δx ≥ (1.055 x 10^-34) / (6.006 x 10^-32) ≈ 1.75 x 10^-3 m
Final Result
Thus, the uncertainty in the position of the electron is approximately:
Δx ≈ 1.75 x 10^-3 m or 1.75 mm
This result illustrates the inherent limitations in measuring the position of subatomic particles like electrons, highlighting the fundamental principles of quantum mechanics. The larger the uncertainty in momentum, the greater the uncertainty in position, which is a core concept in understanding the behavior of particles at the quantum level.