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. `2.36xx10^(-28)m`
B. `5.25xx10^(-28)m`
C. `2.27xx10^(-30)m`
D. `5.27xx10^(-30)m`

Explanation

To find the uncertainty in position (Δx) of an electron given the uncertainty in momentum (Δp), we can use Heisenberg's uncertainty principle, which states:

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

Where:
- \( \Delta x \) is the uncertainty in position,
- \( \Delta p \) is the uncertainty in momentum,
- \( h \) is Planck's constant (\( 6.62 \times 10^{-34} \, \text{kg m}^2/\text{s} \)),
- \( \pi \) is approximately \( 3.14 \).

### Step-by-Step Solution:

1. **Identify Given Values**:
- Uncertainty in momentum, \( \Delta p = 1 \times 10^{-5} \, \text{kg m/s} \)
- Planck's constant, \( h = 6.62 \times 10^{-34} \, \text{kg m}^2/\text{s} \)
- Value of \( \pi \approx 3.14 \)

2. **Use the Uncertainty Principle Formula**:
Rearranging the formula to find \( \Delta x \):
\[ \Delta x = \frac{h}{4\pi \Delta p} \]

3. **Substitute the Values**:
\[ \Delta x = \frac{6.62 \times 10^{-34}}{4 \times 3.14 \times (1 \times 10^{-5})} \]

4. **Calculate the Denominator**:
- First, calculate \( 4 \times 3.14 \):
\[ 4 \times 3.14 = 12.56 \]
- Now calculate \( 12.56 \times (1 \times 10^{-5}) \):
\[ 12.56 \times 10^{-5} = 1.256 \times 10^{-4} \]

5. **Calculate \( \Delta x \)**:
Now substitute back into the equation:
\[ \Delta x = \frac{6.62 \times 10^{-34}}{1.256 \times 10^{-4}} \]
- Performing the division:
\[ \Delta x = 5.27 \times 10^{-30} \, \text{m} \]

### Final Answer:
The uncertainty in position \( \Delta x \) is:
\[ \Delta x = 5.27 \times 10^{-30} \, \text{m} \]