Uncertainty in the position of an electron mass ` (9. 1 xx 10^(31) kg)` moving with a velocity ` 300 ms^(-1)` accurate uptp ` 0.001 %` will be :

Options

A. ` 19 . 2 xx 10^(-2) m`
B. ` 5. 75 xx 10^(-2) m`
C. `1. 92 xx 10^(-5) m`
D. `3. 584 xx 10^(-2) m`

Explanation

To solve the problem of finding the uncertainty in the position of an electron moving with a given velocity, we will use the Heisenberg Uncertainty Principle. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- Mass of the electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Velocity of the electron, \( v = 300 \, \text{m/s} \)
- Accuracy of the velocity, \( 0.001\% \)

### Step 2: Calculate the Uncertainty in Velocity (\( \Delta v \))
The uncertainty in velocity can be calculated as:
\[ \Delta v = \frac{0.001}{100} \times v = \frac{0.001}{100} \times 300 \, \text{m/s} \]
Calculating this gives:
\[ \Delta v = 0.003 \, \text{m/s} \]

### Step 3: Apply the Heisenberg Uncertainty Principle
According to the Heisenberg Uncertainty Principle:
\[ \Delta p \cdot \Delta x \geq \frac{h}{4\pi} \]
Where:
- \( \Delta p \) is the uncertainty in momentum.
- \( h \) is Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{Js} \).

### Step 4: Calculate the Uncertainty in Momentum (\( \Delta p \))
The momentum \( p \) is given by:
\[ p = m \cdot v \]
Thus, the uncertainty in momentum is:
\[ \Delta p = m \cdot \Delta v \]
Substituting the values:
\[ \Delta p = 9.1 \times 10^{-31} \, \text{kg} \cdot 0.003 \, \text{m/s} = 2.73 \times 10^{-33} \, \text{kg m/s} \]

### Step 5: Substitute into the Uncertainty Principle Equation
Now substituting \( \Delta p \) into the Heisenberg equation:
\[ \Delta x \geq \frac{h}{4\pi \Delta p} \]
Substituting the values:
\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{4 \cdot \pi \cdot 2.73 \times 10^{-33}} \]

### Step 6: Calculate \( \Delta x \)
Calculating the denominator:
\[ 4 \cdot \pi \cdot 2.73 \times 10^{-33} \approx 3.42 \times 10^{-32} \]
Now substituting this back into the equation for \( \Delta x \):
\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{3.42 \times 10^{-32}} \approx 0.0194 \, \text{m} \]
This can be expressed as:
\[ \Delta x \approx 1.94 \times 10^{-2} \, \text{m} \]

### Final Answer
The uncertainty in the position of the electron is approximately:
\[ \Delta x \approx 1.94 \times 10^{-2} \, \text{m} \]