Calculate the uncertain it in the velocity of a cricket ball of mass 150 g, if uncertainity in its position is of the order of 1 Å.

Answer

To calculate the uncertainty in the velocity of a cricket ball of mass 150 g given the uncertainty in its position is of the order of 1 Å, we will use Heisenberg's Uncertainty Principle.

### Step-by-Step Solution:

1. **Understand Heisenberg's Uncertainty Principle**:
The principle states that the product of the uncertainty in position (Δx) and the uncertainty in momentum (Δp) is greater than or equal to a constant:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
where \( h \) is Planck's constant.

2. **Identify the given values**:
- Mass of the cricket ball, \( m = 150 \, \text{g} = 0.150 \, \text{kg} \) (since 1 g = 0.001 kg).
- Uncertainty in position, \( \Delta x = 1 \, \text{Å} = 1 \times 10^{-10} \, \text{m} \).
- Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{Js} \).

3. **Calculate the uncertainty in momentum (Δp)**:
The uncertainty in momentum is given by:
\[ \Delta p = m \cdot \Delta v \]
where \( \Delta v \) is the uncertainty in velocity.

4. **Substitute Δp into the uncertainty principle**:
Replacing Δp in the uncertainty principle formula:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

5. **Rearrange to solve for Δv**:
\[ \Delta v \geq \frac{h}{4\pi \Delta x \cdot m} \]

6. **Substitute the known values**:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4 \cdot \pi \cdot (1 \times 10^{-10}) \cdot (0.150)} \]

7. **Calculate the denominator**:
\[ 4 \cdot \pi \cdot (1 \times 10^{-10}) \cdot (0.150) \approx 1.884 \times 10^{-10} \]

8. **Calculate Δv**:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{1.884 \times 10^{-10}} \approx 3.51 \times 10^{-24} \, \text{m/s} \]

### Final Result:
The uncertainty in the velocity of the cricket ball is approximately \( 3.51 \times 10^{-24} \, \text{m/s} \).