0.5 gm particle has uncertainty
of `2xx10^(-5)` m find the uncertainty in its velocity`(m//s)`

Options

A. `3.0xx10^(33)`
B. `5xx10^(-27)`
C. `4xx10^(-19)`
D. `4xx10^(-10)`

Explanation

To find the uncertainty in the velocity of a particle with a given uncertainty in position, we can use Heisenberg's uncertainty principle. Here’s a step-by-step solution:

### Step 1: Understand Heisenberg's Uncertainty Principle
Heisenberg's uncertainty principle states that:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
where:
- \(\Delta x\) = uncertainty in position
- \(\Delta p\) = uncertainty in momentum
- \(h\) = Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\))

### Step 2: Relate Momentum to Velocity
Momentum (\(p\)) is given by the product of mass (\(m\)) and velocity (\(v\)):
\[ p = mv \]
Thus, the uncertainty in momentum can be expressed as:
\[ \Delta p = m \cdot \Delta v \]
where \(\Delta v\) is the uncertainty in velocity.

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

### Step 4: Rearrange to Solve for \(\Delta v\)
Rearranging the equation to solve for \(\Delta v\):
\[ \Delta v = \frac{h}{4\pi m \Delta x} \]

### Step 5: Insert Given Values
Now we can substitute the known values:
- \(h = 6.626 \times 10^{-34} \, \text{Js}\)
- \(m = 0.5 \, \text{g} = 0.5 \times 10^{-3} \, \text{kg}\)
- \(\Delta x = 2 \times 10^{-5} \, \text{m}\)

Substituting these values into the equation:
\[ \Delta v = \frac{6.626 \times 10^{-34}}{4 \cdot 3.14 \cdot (0.5 \times 10^{-3}) \cdot (2 \times 10^{-5})} \]

### Step 6: Calculate the Denominator
Calculating the denominator:
\[ 4 \cdot 3.14 \cdot (0.5 \times 10^{-3}) \cdot (2 \times 10^{-5}) = 4 \cdot 3.14 \cdot 0.5 \cdot 2 \times 10^{-8} \]
Calculating the numerical part:
\[ 4 \cdot 3.14 \cdot 0.5 \cdot 2 = 12.56 \]
Thus:
\[ 12.56 \times 10^{-8} = 1.256 \times 10^{-7} \]

### Step 7: Final Calculation
Now substituting back:
\[ \Delta v = \frac{6.626 \times 10^{-34}}{1.256 \times 10^{-7}} \approx 5.28 \times 10^{-27} \, \text{m/s} \]

### Step 8: Round the Result
Rounding to significant figures:
\[ \Delta v \approx 5 \times 10^{-27} \, \text{m/s} \]

### Final Answer
The uncertainty in the velocity of the particle is approximately:
\[ \Delta v \approx 5 \times 10^{-27} \, \text{m/s} \]