We are given the following information:

• The mass of the particle, $m = 2 \times 10^{-9} \, \text{kg}$
• The speed of the particle, $v = 3.00 \, \text{m/s}$
• The uncertainty in speed, $\Delta v = 0.4\%$ of the speed $v$.

Step 1: Calculate the uncertainty in speed

First, we calculate the absolute uncertainty in the speed, $\Delta v$: $\Delta v = 0.4\% \times 3.00 \, \text{m/s} = 0.004 \times 3.00 = 0.012 \, \text{m/s}$

Step 2: Use Heisenberg's Uncertainty Principle

To find the uncertainty in position, $\Delta x$, we use Heisenberg's Uncertainty Principle, which is given by the inequality: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$ where $\Delta p$ is the uncertainty in momentum, and $h$ is Planck's constant ($h = 6.626 \times 10^{-34} \, \text{Js}$).

Step 3: Calculate uncertainty in momentum

The uncertainty in momentum, $\Delta p$, is related to the uncertainty in speed $\Delta v$ by the equation: $\Delta p = m \cdot \Delta v$ Substituting the known values: $\Delta p = (2 \times 10^{-9} \, \text{kg}) \times (0.012 \, \text{m/s}) = 2.4 \times 10^{-11} \, \text{kg} \cdot \text{m/s}$

Step 4: Calculate uncertainty in position

Now, using Heisenberg's Uncertainty Principle to find $\Delta x$: $\Delta x \geq \frac{h}{4\pi \Delta p}$ Substitute the known values: $\Delta x \geq \frac{6.626 \times 10^{-34}}{4\pi \times 2.4 \times 10^{-11}} = \frac{6.626 \times 10^{-34}}{3.016 \times 10^{-10}} \approx 2.2 \times 10^{-24} \, \text{m}$

Final Answer

Thus, the uncertainty in the position is approximately: $\Delta x \approx 2.20 \times 10^{-24} \, \text{m}$

Would you like further clarification on any part of this?

Here are 5 related questions to consider:

1. How does Heisenberg's Uncertainty Principle apply to macroscopic objects?
2. What is the significance of Planck's constant in quantum mechanics?
3. How does the uncertainty in speed affect the uncertainty in position?
4. Can the uncertainty in momentum be reduced to zero? Why or why not?
5. How would the uncertainty change if the particle's mass were larger?

Tip: In quantum mechanics, reducing the uncertainty in one observable (like position) increases the uncertainty in its conjugate observable (like momentum).