Uncertinity in position of a 0.25g particle is `10^(-5)m` then the uncertinity in its velecity will be

Options

A. `1.2 xx 10^34`
B. `2.1 xx 10^(-26)`
C. `1.6 xx 10^(-20)`
D. `1.7 xx 10^(-9)`

Answer

To find the uncertainty in the velocity of a particle given its uncertainty in position, we will use the Heisenberg Uncertainty Principle. The principle states that:

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

Where:
- \(\Delta x\) is the uncertainty in position,
- \(\Delta p\) is the uncertainty in momentum,
- \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)).

### Step-by-Step Solution:

1. **Identify Given Values**:
- Uncertainty in position, \(\Delta x = 10^{-5} \, \text{m}\)
- Mass of the particle, \(m = 0.25 \, \text{g} = 0.25 \times 10^{-3} \, \text{kg}\)

2. **Convert Mass to Kilograms**:
- Since the mass is given in grams, we convert it to kilograms:
\[ m = 0.25 \, \text{g} = 0.25 \times 10^{-3} \, \text{kg} \]

3. **Express Uncertainty in Momentum**:
- The uncertainty in momentum \(\Delta p\) can be expressed as:
\[ \Delta p = m \cdot \Delta v \]
where \(\Delta v\) is the uncertainty in velocity.

4. **Substitute into the Uncertainty Principle**:
- Substitute \(\Delta p\) in the Heisenberg Uncertainty Principle:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

5. **Rearranging the Equation**:
- Rearranging to find \(\Delta v\):
\[ \Delta v \geq \frac{h}{4\pi \Delta x m} \]

6. **Substituting the Known Values**:
- Substitute the known values into the equation:
\[ \Delta v \geq \frac{6.63 \times 10^{-34}}{4 \cdot 3.14 \cdot (10^{-5}) \cdot (0.25 \times 10^{-3})} \]

7. **Calculating**:
- Calculate the denominator:
\[ 4 \cdot 3.14 \cdot 10^{-5} \cdot (0.25 \times 10^{-3}) = 4 \cdot 3.14 \cdot 0.25 \cdot 10^{-8} = 3.14 \times 10^{-8} \]
- Now calculate \(\Delta v\):
\[ \Delta v \geq \frac{6.63 \times 10^{-34}}{3.14 \times 10^{-8}} \approx 2.11 \times 10^{-26} \, \text{m/s} \]

8. **Final Result**:
- Therefore, the uncertainty in velocity \(\Delta v\) is approximately:
\[ \Delta v \approx 2.1 \times 10^{-26} \, \text{m/s} \]