An electron moving near an atomic nucleus has a speed of `6xx10^(6) +- 1% m//s`. What is the uncertainty in its position?

Explanation

To solve the problem of finding the uncertainty in the position of an electron moving near an atomic nucleus, we will follow these steps:

### Step 1: Calculate the uncertainty in speed (ΔV)
The speed of the electron is given as \(6 \times 10^6 \, \text{m/s} \pm 1\%\). To find the uncertainty in speed, we calculate:

\[ \Delta V = \text{Speed} \times \text{Percentage Error} \]

Given that the percentage error is \(1\%\) or \(0.01\):

\[ \Delta V = 6 \times 10^6 \, \text{m/s} \times 0.01 = 6 \times 10^4 \, \text{m/s} \]

### Step 2: Use the Heisenberg Uncertainty Principle
According to the Heisenberg Uncertainty Principle, the uncertainty in position (Δx) can be calculated using the formula:

\[ \Delta x \geq \frac{h}{4 \pi m \Delta V} \]

Where:
- \(h\) is Planck's constant, approximately \(6.626 \times 10^{-34} \, \text{Js}\)
- \(m\) is the mass of the electron, approximately \(9.11 \times 10^{-31} \, \text{kg}\)
- \(\Delta V\) is the uncertainty in speed calculated in Step 1.

### Step 3: Substitute the values into the formula
Now, we substitute the values into the formula:

\[ \Delta x \geq \frac{6.626 \times 10^{-34} \, \text{Js}}{4 \pi (9.11 \times 10^{-31} \, \text{kg})(6 \times 10^4 \, \text{m/s})} \]

### Step 4: Calculate the denominator
First, calculate the denominator:

\[ 4 \pi (9.11 \times 10^{-31} \, \text{kg})(6 \times 10^4 \, \text{m/s}) \approx 4 \times 3.14 \times 9.11 \times 10^{-31} \times 6 \times 10^4 \]

Calculating this gives:

\[ \approx 6.94 \times 10^{-26} \, \text{kg m/s} \]

### Step 5: Calculate Δx
Now, substituting back into the equation for Δx:

\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{6.94 \times 10^{-26}} \approx 9.55 \times 10^{-9} \, \text{m} \]

### Step 6: Final result
Thus, the uncertainty in the position of the electron is:

\[ \Delta x \geq 1 \times 10^{-9} \, \text{m} \]

### Summary
The uncertainty in the position of the electron is approximately \(1 \times 10^{-9} \, \text{m}\).

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