Question

German physicist Werner Heisenberg related the uncertainty of an object\'s position (Δx) to the uncertainty in its velocity Δv. The mass of an electron is 9.11× 10–31 kg. What is the uncertainty in the position of an electron moving at 2.00 × 106 m/s with an uncertainty of Δv = 0.01 × 106 m/s?

Answer 1

Heisenberg's Uncertainty Principle gives a relationship between the standard deviation of an object's position and its momentum.

$\Delta p \cdot \Delta x = h / (4 \pi)$ where

• $\Delta p$ the standard deviation of the object's momentum,
• $\Delta x$ the standard deviation of the object's position, and
• $h \approx 6.63 \times 10^{-34} \; \text{J} \cdot \text{s}$ the Planck's constant.

By definition, the momentum of the electron equals the product of its mass and velocity.

$p = m\cdot v$

Assuming that measurement of the mass of the electron $m$ is accurate. It is assumed to be a coefficient of constant value. The standard deviation in the electron's velocity is thus directly related to that of its mass. That is:

$\Delta p = m \cdot \Delta v$

$\Delta v = 0.01 \times 10^{6} \;\text{m}\cdot \text{s}^{-1}$ from the question;

$\Delta p = m\cdot v \\ \phantom{\Delta p} = 0.01 \times 10^{6} \; \text{m} \cdot \text{s}^{-1} \times 9.11 \times 10^{-31} \; \text{kg}\\\phantom{\Delta p} = 9.11 \times 10^{-27} \; \text{kg} \cdot \text{m}\cdot \text{s}^{-1}$

Convert the unit of the Planck's constant to base SI units (kg, m, s, etc.) if it was provided in derived units such as joules. Doing so would allow for a dimension analysis on the accuracy of the result.

$h = 6.63 \times 10^{-34} \; \text{J} \cdot \text{s}\\\phantom{h} = 6.63 \times 10^{-34} \; (\text{N}\cdot \text{m}) \cdot \text{s} \\\phantom{h} = 6.63 \times 10^{-34} \; ((\text{kg} \cdot \text{m}\cdot \text{s}^{-2}) \cdot \text{m}) \cdot \text{s}\\\phantom{h} = 6.63 \times 10^{-34} \; \text{kg} \cdot \text{m}^{2} \cdot \text{s}^{-1}$

Apply the Uncertainty Principle:

$\Delta x = h/ (4 \pi \cdot \Delta p)\\\phantom{\Delta x} = 6.63 \times 10^{-34} \; \text{kg} \cdot \text{m}^{2} \cdot \text{s}^{-1} / (4 \pi \cdot 9.11\times 10^{-27} \; \text{kg} \cdot \text{m}\cdot \text{s}^{-1})\\\phantom{\Delta x} = 5.79 \times 10^{-9} \; \text{m}$.

Dimensional analysis:

$\Delta x$ resembles the standard deviation of a position measurement. It is expected to have a unit of meter, which is the same as that of position.