The uncertainity in the momentum of an electron is `1.0xx10^(-5)"kg m s"^(-1)`. The uncertainity in its position will be: `(h=6.626xx10^(-34)Js)`

Options

A. `1.05 xx 10^(-28)m`
B. `1.05 xx 10^(-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. This principle states that the product of the uncertainties in position (Δx) and momentum (Δp) is equal to or greater than a constant value, which 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 (6.626 x 10^(-34) Js)

### Step-by-step Solution:

1. **Identify the given values:**
- Uncertainty in momentum (Δp) = \( 1.0 \times 10^{-5} \, \text{kg m/s} \)
- Planck's constant (h) = \( 6.626 \times 10^{-34} \, \text{Js} \)

2. **Use the uncertainty principle formula:**
Rearranging the formula to find Δx:
\[ \Delta x = \frac{h}{4\pi \Delta p} \]

3. **Substitute the known values into the formula:**
\[ \Delta x = \frac{6.626 \times 10^{-34}}{4 \times \pi \times 1.0 \times 10^{-5}} \]

4. **Calculate the denominator:**
- First, calculate \( 4 \times \pi \):
\[ 4 \times \pi \approx 4 \times 3.14 \approx 12.56 \]
- Now, calculate \( 4 \times \pi \times 1.0 \times 10^{-5} \):
\[ 12.56 \times 10^{-5} = 1.256 \times 10^{-4} \]

5. **Now, compute Δx:**
\[ \Delta x = \frac{6.626 \times 10^{-34}}{1.256 \times 10^{-4}} \approx 5.27 \times 10^{-30} \, \text{m} \]

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