An atom has mass of 0.02 kg and uncertainty in its velocity is 9.218×`10^(−6)`m/s then uncertainty in position is (h=6.626×`10^(−34)`Js)

Options

A. 2.86×`10^(−23)`m
B. 2.86×`10^(−28)`m
C. 1.86×`10^(−28)`m
D. None of these

Explanation

To find the uncertainty in position (Δx) of the atom, we will use the Heisenberg uncertainty principle, which states:

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

Where:
- Δx = uncertainty in position
- m = mass of the atom
- Δv = uncertainty in velocity
- h = Planck's constant

Given:
- Mass of the atom (m) = 0.02 kg
- Uncertainty in velocity (Δv) = \(9.218 \times 10^{-6}\) m/s
- Planck's constant (h) = \(6.626 \times 10^{-34}\) Js

### Step 1: Rearranging the Uncertainty Principle
We need to isolate Δx:

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

### Step 2: Substitute the Known Values
Now we substitute the known values into the equation:

\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{4 \cdot \pi \cdot 0.02 \cdot 9.218 \times 10^{-6}} \]

### Step 3: Calculate the Denominator
First, calculate the denominator:

\[ 4 \cdot \pi \cdot 0.02 \cdot 9.218 \times 10^{-6} \]

Using \(\pi \approx 3.14\):

\[ 4 \cdot 3.14 \cdot 0.02 \cdot 9.218 \times 10^{-6} \approx 2.33 \times 10^{-7} \]

### Step 4: Calculate Δx
Now, we can calculate Δx:

\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{2.33 \times 10^{-7}} \approx 2.85 \times 10^{-27} \text{ m} \]

### Step 5: Final Value
Thus, the uncertainty in position (Δx) is approximately:

\[ \Delta x \approx 2.86 \times 10^{-28} \text{ m} \]

### Conclusion
The uncertainty in position of the atom is \(2.86 \times 10^{-28}\) meters.

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