The uncertainty in the momentum of a particle is `3.3 xx10^(-2) kg ms^(-1)` the uncertainty in its position will be

Options

A. `1.6 xx 10^(-33)`m
B. `1.6 xx 10^(-32)`m
C. `1.6 xx 10^(-30)`m
D. `1.6 xx 10^(-29)`m

Answer

To find the uncertainty in the position of a particle 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, approximately \(6.626 \times 10^{-34} \, \text{kg m}^2/\text{s}\).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Uncertainty in momentum, \(\Delta p = 3.3 \times 10^{-2} \, \text{kg m/s}\).
- Planck's constant, \(h = 6.626 \times 10^{-34} \, \text{kg m}^2/\text{s}\).

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

3. **Substitute the Values**:
Substitute \(h\) and \(\Delta p\) into the equation:
\[ \Delta x = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 3.3 \times 10^{-2}} \]

4. **Calculate the Denominator**:
First, calculate the denominator:
\[ 4 \times 3.14 \times 3.3 \times 10^{-2} = 4 \times 3.14 \times 0.033 = 0.41568 \approx 0.416 \]

5. **Calculate \(\Delta x\)**:
Now, calculate \(\Delta x\):
\[ \Delta x = \frac{6.626 \times 10^{-34}}{0.416} \approx 1.59 \times 10^{-33} \, \text{m} \]

6. **Round the Result**:
Rounding \(1.59 \times 10^{-33}\) gives approximately:
\[ \Delta x \approx 1.6 \times 10^{-33} \, \text{m} \]

7. **Conclusion**:
Therefore, the uncertainty in the position of the particle is:
\[ \Delta x \approx 1.6 \times 10^{-33} \, \text{m} \]

### Final Answer:
The uncertainty in its position is \(1.6 \times 10^{-33} \, \text{m}\).