A cricket ball weighing 100 g is located within 1 nm. What is the uncertainty in the velocity ?

Explanation

To find the uncertainty in the velocity of a cricket ball weighing 100 g located within 1 nm, we will use the Heisenberg Uncertainty Principle. Here’s a step-by-step solution:

### Step 1: Understand the Heisenberg Uncertainty Principle
The Heisenberg Uncertainty 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.

### Step 2: Identify the given values
- Mass of the cricket ball (m) = 100 g = 0.1 kg (convert grams to kilograms)
- Uncertainty in position (Δx) = 1 nm = \( 1 \times 10^{-9} \) m (convert nanometers to meters)

### Step 3: Relate momentum to velocity
Momentum (p) is given by the product of mass and velocity:
\[ p = mv \]
Thus, the uncertainty in momentum (Δp) can be expressed as:
\[ \Delta p = m \cdot \Delta v \]

### Step 4: Substitute Δp in the uncertainty principle
Substituting Δp in the Heisenberg Uncertainty Principle:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

### Step 5: Solve for Δv
Rearranging the equation to find Δv:
\[ \Delta v \geq \frac{h}{4\pi m \Delta x} \]

### Step 6: Substitute the known values
Now, substituting the known values:
- Planck's constant \( h = 6.63 \times 10^{-34} \) J·s
- \( m = 0.1 \) kg
- \( \Delta x = 1 \times 10^{-9} \) m

\[ \Delta v \geq \frac{6.63 \times 10^{-34}}{4 \cdot \pi \cdot 0.1 \cdot 1 \times 10^{-9}} \]

### Step 7: Calculate Δv
Calculating the right-hand side:
\[ \Delta v \geq \frac{6.63 \times 10^{-34}}{4 \cdot 3.14 \cdot 0.1 \cdot 1 \times 10^{-9}} \]
\[ \Delta v \geq \frac{6.63 \times 10^{-34}}{1.25664 \times 10^{-10}} \approx 5.27 \times 10^{-24} \text{ m/s} \]

### Final Answer
Thus, the uncertainty in the velocity of the cricket ball is approximately:
\[ \Delta v \geq 5.27 \times 10^{-25} \text{ m/s} \]

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