The uncertainty in the momentum of an electron is `1.2x10^-5 kg ms^-1`.The uncertainty in its position will be

Options

A. `1.50x10^-26m
B. `1.05x 10^-26m`
C. `5.27x10^-30 m`
D. `5.25x10^-28 m`

Answer

To find the uncertainty in the position of an electron given the uncertainty in its momentum, we will 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, approximately \(6.626 \times 10^{-34} \, \text{Js}\).

### Step-by-Step Solution:

1. **Identify the given values**:
- Uncertainty in momentum, \(\Delta p = 1.2 \times 10^{-5} \, \text{kg m/s}\).
- Planck's constant, \(h = 6.626 \times 10^{-34} \, \text{Js}\).

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

3. **Substitute the values into the equation**:
- First, calculate \(4\pi\):
\[ 4\pi \approx 4 \times 3.14 = 12.56 \]
- Now substitute the values into the equation:
\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{12.56 \times (1.2 \times 10^{-5})} \]

4. **Calculate the denominator**:
\[ 12.56 \times (1.2 \times 10^{-5}) \approx 1.5072 \times 10^{-4} \]

5. **Calculate \(\Delta x\)**:
\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{1.5072 \times 10^{-4}} \approx 4.4 \times 10^{-30} \, \text{m} \]

### Final Answer:
The uncertainty in the position of the electron is approximately \(4.4 \times 10^{-30} \, \text{m}\).

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