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 solve the problem of finding the uncertainty in the velocity of a particle given the uncertainty in its position, we can 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: Relate momentum to mass and velocity
The momentum \(p\) of an object is given by:
\[ p = m \cdot v \]
where \(m\) is the mass and \(v\) is the velocity. Therefore, the uncertainty in momentum (Δp) can be expressed as:
\[ \Delta p = m \cdot \Delta v \]
where Δv is the uncertainty in velocity.

### Step 3: Substitute into the uncertainty principle
Substituting Δp into the uncertainty principle gives:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

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

### Step 5: Insert known values
We know:
- \(h = 6.63 \times 10^{-34} \, \text{J s}\)
- \(\Delta x = 10^{-5} \, \text{m}\)
- The mass of the particle is \(0.25 \, \text{g} = 0.25 \times 10^{-3} \, \text{kg} = 2.5 \times 10^{-4} \, \text{kg}\)

Now substituting these values into the equation:
\[ \Delta v \geq \frac{6.63 \times 10^{-34}}{4\pi (2.5 \times 10^{-4}) (10^{-5})} \]

### Step 6: Calculate the right-hand side
Calculating the denominator:
\[ 4\pi (2.5 \times 10^{-4}) (10^{-5}) = 4\pi \times 2.5 \times 10^{-9} \approx 3.14 \times 10^{-8} \, \text{(using } \pi \approx 3.14\text{)} \]

Now substituting this back:
\[ \Delta v \geq \frac{6.63 \times 10^{-34}}{3.14 \times 10^{-8}} \approx 2.11 \times 10^{-26} \, \text{m/s} \]

### Conclusion
Thus, the uncertainty in the velocity of the particle is approximately:
\[ \Delta v \geq 2.1 \times 10^{-26} \, \text{m/s} \]

### Final Answer
The uncertainty in the velocity of the particle is \(2.1 \times 10^{-26} \, \text{m/s}\).

---