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 find the uncertainty in the position of an electron given the uncertainty in its momentum, we can use the Heisenberg Uncertainty Principle. The principle states that:

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

Where:
- \(\Delta x\) is the uncertainty in position,
- \(\Delta p\) is the uncertainty in momentum,
- \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{J s}\)).

### 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, \(h = 6.63 \times 10^{-34} \, \text{J s}\)

2. **Rearrange the uncertainty principle equation to solve for \(\Delta x\)**:
\[ \Delta x = \frac{h}{4\pi \Delta p} \]

3. **Substitute the known values into the equation**:
- First, calculate \(4\pi\):
\[ 4\pi \approx 4 \times \frac{22}{7} \approx 12.57 \]
- Now substitute \(h\) and \(\Delta p\):
\[ \Delta x = \frac{6.63 \times 10^{-34}}{12.57 \times (1 \times 10^{-5})} \]

4. **Calculate the denominator**:
\[ 12.57 \times 1 \times 10^{-5} \approx 1.257 \times 10^{-4} \]

5. **Calculate \(\Delta x\)**:
\[ \Delta x = \frac{6.63 \times 10^{-34}}{1.257 \times 10^{-4}} \approx 5.27 \times 10^{-30} \, \text{m} \]

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