An electron has a speed of 500m `s^(-1)` with uncertainty of 0.02% . What is the uncertainty in locating its position ?

Explanation

To find the uncertainty in locating the position of an electron with a given speed and uncertainty, we can use Heisenberg's uncertainty principle. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Speed of the electron, \( v = 500 \, \text{m/s} \)
- Uncertainty in speed, \( \Delta v = 0.02\% \)

### Step 2: Calculate the uncertainty in speed (\( \Delta v \))
To find the uncertainty in speed, we calculate \( 0.02\% \) of \( 500 \, \text{m/s} \):
\[ \Delta v = 0.02\% \times 500 = \frac{0.02}{100} \times 500 = 0.1 \, \text{m/s} \]

### Step 3: Use Heisenberg's uncertainty principle
Heisenberg's uncertainty principle is given by:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
where:
- \( \Delta x \) is the uncertainty in position,
- \( \Delta p \) is the uncertainty in momentum,
- \( h \) is Planck's constant, approximately \( 6.63 \times 10^{-34} \, \text{Js} \).

### Step 4: Express uncertainty in momentum (\( \Delta p \))
The uncertainty in momentum can be expressed as:
\[ \Delta p = m \cdot \Delta v \]
where \( m \) is the mass of the electron, approximately \( 9.1 \times 10^{-31} \, \text{kg} \).

### Step 5: Calculate \( \Delta p \)
Substituting the values:
\[ \Delta p = 9.1 \times 10^{-31} \, \text{kg} \cdot 0.1 \, \text{m/s} = 9.1 \times 10^{-32} \, \text{kg m/s} \]

### Step 6: Substitute into the uncertainty principle formula
Now, we substitute \( \Delta p \) into the Heisenberg's uncertainty principle:
\[ \Delta x \cdot (9.1 \times 10^{-32}) \geq \frac{6.63 \times 10^{-34}}{4\pi} \]

### Step 7: Calculate \( \frac{h}{4\pi} \)
Calculating \( \frac{h}{4\pi} \):
\[ \frac{6.63 \times 10^{-34}}{4 \times 3.14} \approx \frac{6.63 \times 10^{-34}}{12.56} \approx 5.28 \times 10^{-35} \, \text{m}^2 \text{kg/s} \]

### Step 8: Solve for \( \Delta x \)
Now, we can solve for \( \Delta x \):
\[ \Delta x \geq \frac{5.28 \times 10^{-35}}{9.1 \times 10^{-32}} \approx 5.8 \times 10^{-4} \, \text{m} \]

### Step 9: Final result
Thus, the uncertainty in locating the position of the electron is approximately:
\[ \Delta x \approx 0.58 \times 10^{-3} \, \text{m} = 0.57 \, \text{mm} \]

### Summary
The uncertainty in locating the position of the electron is \( 0.57 \, \text{mm} \).

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