The uncertainty in momentum of an electron is `1 xx 10^-5 kg - m//s`. The uncertainty in its position will be `(h = 6.62 xx 10^-34 kg = m^2//s)`.

Options

A. `1.05xx10^(-28)m`
B. `1.05xx10^(-26)m`
C. `5.27xx10^(-30)m`
D. `5.25xx10^(-28)m`

Explanation

To solve the problem, we will use the Heisenberg Uncertainty Principle, which states that the product of the uncertainties in position (Δx) and momentum (Δp) of a particle is given by the equation:

\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]

Where:
- \( \Delta x \) = uncertainty in position
- \( \Delta p \) = uncertainty in momentum
- \( h \) = Planck's constant (given as \( 6.62 \times 10^{-34} \, \text{kg m}^2/\text{s} \))

### Step-by-Step Solution:

1. **Identify the given values**:
- Uncertainty in momentum, \( \Delta p = 1 \times 10^{-5} \, \text{kg m/s} \)
- Planck's constant, \( h = 6.62 \times 10^{-34} \, \text{kg m}^2/\text{s} \)

2. **Use the Heisenberg Uncertainty Principle**:
- Rearranging the equation to solve for \( \Delta x \):
\[ \Delta x = \frac{h}{4\pi \Delta p} \]

3. **Substitute the values into the equation**:
- First, calculate \( 4\pi \):
\[ 4\pi \approx 4 \times 3.14 \approx 12.56 \]
- Now substitute \( h \) and \( \Delta p \):
\[ \Delta x = \frac{6.62 \times 10^{-34}}{12.56 \times 1 \times 10^{-5}} \]

4. **Calculate the denominator**:
\[ 12.56 \times 1 \times 10^{-5} = 1.256 \times 10^{-4} \]

5. **Calculate \( \Delta x \)**:
\[ \Delta x = \frac{6.62 \times 10^{-34}}{1.256 \times 10^{-4}} \approx 5.27 \times 10^{-30} \, \text{m} \]

### Final Answer:
The uncertainty in the position of the electron is approximately:
\[ \Delta x \approx 5.27 \times 10^{-30} \, \text{m} \]