An atom has mass of 0.01 kg and uncertainty in its velocity is 7.318×`10^(−6)`m/s then uncertainty in position is (h=6.626×`10^(−34)`Js)

Options

A. 2.86×`10^(−23)`m
B. 7.2×`10^(−28)`m
C. 6.86×`10^(−23)`m
D. None

Explanation

To find the uncertainty in position (Δx) of an atom given its mass (m), uncertainty in velocity (Δv), and Planck's constant (h), we can use the Heisenberg Uncertainty Principle, which states:

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

### Step-by-step Solution:

1. **Identify the given values:**
- Mass of the atom, \( m = 0.01 \, \text{kg} \)
- Uncertainty in velocity, \( \Delta v = 7.318 \times 10^{-6} \, \text{m/s} \)
- Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{Js} \)

2. **Substitute the known values into the uncertainty principle equation:**
\[ \Delta x \cdot (0.01 \, \text{kg}) \cdot (7.318 \times 10^{-6} \, \text{m/s}) \geq \frac{6.626 \times 10^{-34} \, \text{Js}}{4\pi} \]

3. **Calculate \( 4\pi \):**
\[ 4\pi \approx 4 \times 3.14 = 12.56 \]

4. **Calculate \( \frac{h}{4\pi} \):**
\[ \frac{6.626 \times 10^{-34}}{12.56} \approx 5.280 \times 10^{-35} \, \text{Js} \]

5. **Rearranging the equation to solve for \( \Delta x \):**
\[ \Delta x \geq \frac{5.280 \times 10^{-35}}{0.01 \times 7.318 \times 10^{-6}} \]

6. **Calculate the denominator:**
\[ 0.01 \times 7.318 \times 10^{-6} = 7.318 \times 10^{-8} \]

7. **Now, calculate \( \Delta x \):**
\[ \Delta x \geq \frac{5.280 \times 10^{-35}}{7.318 \times 10^{-8}} \approx 7.20 \times 10^{-28} \, \text{m} \]

### Final Result:
The uncertainty in position \( \Delta x \) is approximately \( 7.20 \times 10^{-28} \, \text{m} \).