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. `1.05 xx 10^-28 m`
B. `1.05 xx 10^-26 m`
C. `5.27 xx 10^-30 m`
D. `5.25 xx 10^-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, which states:

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

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

### 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. **Apply the Heisenberg Uncertainty Principle:**
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]

3. **Rearrange the equation to solve for \(\Delta x\):**
\[ \Delta x \geq \frac{h}{4\pi \Delta p} \]

4. **Substitute the known values into the equation:**
\[ \Delta x \geq \frac{6.62 \times 10^{-34}}{4 \times 3.14 \times (1 \times 10^{-5})} \]

5. **Calculate the denominator:**
\[ 4 \times 3.14 \times (1 \times 10^{-5}) = 12.56 \times 10^{-5} \]

6. **Now compute \(\Delta x\):**
\[ \Delta x \geq \frac{6.62 \times 10^{-34}}{12.56 \times 10^{-5}} = \frac{6.62}{12.56} \times 10^{-34 + 5} = \frac{6.62}{12.56} \times 10^{-29} \]

7. **Calculate the fraction:**
\[ \frac{6.62}{12.56} \approx 0.527 \]

8. **Final calculation for \(\Delta x\):**
\[ \Delta x \geq 0.527 \times 10^{-29} \, \text{m} = 5.27 \times 10^{-30} \, \text{m} \]

### Conclusion:
The uncertainty in the position of the electron is approximately \(5.27 \times 10^{-30} \, \text{m}\).

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