The mass of a particle is `10^(-10)g` and its radius is `2xx10^(-4)cm`. If its velocity is `10^(-6)cm sec^(-1)` with `0.0001%` uncertainty in measurement, the uncertainty in its position is `:`

Options

A. `5.2xx10^(-8)`m
B. `5.2xx10^(-7)`m
C. `5.2xx10^(-6)`m
D. `5.2xx10^(-9)`

Answer

To find the uncertainty in the position of a particle using the Heisenberg uncertainty principle, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the particle (m) = \(10^{-10} \, \text{g}\)
- Radius (not needed for this calculation)
- Velocity (v) = \(10^{-6} \, \text{cm/s}\)
- Uncertainty in velocity (Δv) = \(0.0001\% \)

2. **Convert Mass to Kilograms:**
- Since \(1 \, \text{g} = 10^{-3} \, \text{kg}\),
\[ m = 10^{-10} \, \text{g} = 10^{-10} \times 10^{-3} \, \text{kg} = 10^{-13} \, \text{kg} \]

3. **Convert Velocity to SI Units:**
- Convert \(10^{-6} \, \text{cm/s}\) to meters per second:
\[ v = 10^{-6} \, \text{cm/s} = 10^{-6} \times 10^{-2} \, \text{m/s} = 10^{-8} \, \text{m/s} \]

4. **Calculate Uncertainty in Velocity (Δv):**
- Calculate \(0.0001\%\) of \(v\):
\[ \Delta v = 0.0001\% \, \text{of} \, 10^{-8} \, \text{m/s} = \frac{0.0001}{100} \times 10^{-8} = 10^{-12} \, \text{m/s} \]

5. **Apply Heisenberg's Uncertainty Principle:**
- The principle states:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
where \(\Delta p = m \cdot \Delta v\).

6. **Calculate Momentum Uncertainty (Δp):**
\[ \Delta p = m \cdot \Delta v = 10^{-13} \, \text{kg} \cdot 10^{-12} \, \text{m/s} = 10^{-25} \, \text{kg m/s} \]

7. **Use Planck's Constant (h):**
- \(h = 6.626 \times 10^{-34} \, \text{J s}\)

8. **Substitute Values into the Uncertainty Equation:**
\[ \Delta x \cdot 10^{-25} \geq \frac{6.626 \times 10^{-34}}{4\pi} \]

9. **Calculate \(\frac{h}{4\pi}\):**
\[ \frac{h}{4\pi} = \frac{6.626 \times 10^{-34}}{4 \times 3.14} \approx \frac{6.626 \times 10^{-34}}{12.56} \approx 5.28 \times 10^{-35} \, \text{m}^2 \text{s} \]

10. **Solve for Uncertainty in Position (Δx):**
\[ \Delta x \geq \frac{5.28 \times 10^{-35}}{10^{-25}} = 5.28 \times 10^{-10} \, \text{m} \]

11. **Convert to Standard Form:**
\[ \Delta x \approx 5.28 \times 10^{-10} \, \text{m} = 5.28 \times 10^{-8} \, \text{cm} \]

### Final Answer:
The uncertainty in the position of the particle is approximately \(5.28 \times 10^{-10} \, \text{m}\).