Werner Heisenberg considered the limits of how precisely we can measure the properties of an electron or other microscopic particle. He determined that there is a fundamental limit to how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse also true. This is summed up in what we now call the Heisenberg uncertainty principle.
The equation si `deltax.delta (mv)ge(h)/(4pi)`
The uncertainty in the position or in the momentum of a marcroscopic object like a baseball is too small to observe. However, the mass of microscopic object such as an electon is small enough for the uncertainty to be relatively large and significant.
If the uncertainty in velocity and position is same, then the uncertainty in momentum will be :

Options

A. `sqrt((hm)/(4pi))`
B. `msqrt((h)/(4pi))`
C. `sqrt((h)/(4pim))`
D. `(1)/(m)sqrt((h)/(4pi))`

Answer

To solve the problem regarding the uncertainty in momentum when the uncertainty in position and velocity are the same, we can follow these steps:

### Step-by-Step Solution

1. **Understand Heisenberg's Uncertainty Principle**:
The principle states that the product of the uncertainties in position (Δx) and momentum (Δp) cannot be smaller than a certain value, which is given by the equation:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]

2. **Relate Momentum to Velocity**:
Momentum (p) is defined as the product of mass (m) and velocity (v):
\[ p = mv \]
Therefore, the uncertainty in momentum (Δp) can be expressed as:
\[ \Delta p = m \cdot \Delta v \]

3. **Set Up the Given Condition**:
According to the problem, the uncertainty in position (Δx) is equal to the uncertainty in velocity (Δv):
\[ \Delta x = \Delta v \]

4. **Substitute Δp in the Uncertainty Principle**:
We can substitute Δp in the uncertainty principle equation:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

5. **Substituting Δx with Δv**:
Since Δx = Δv, we can rewrite the equation as:
\[ \Delta v \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]
This simplifies to:
\[ m \cdot (\Delta v)^2 \geq \frac{h}{4\pi} \]

6. **Solve for Δv**:
Rearranging the equation to find Δv gives:
\[ (\Delta v)^2 \geq \frac{h}{4\pi m} \]
Taking the square root of both sides, we find:
\[ \Delta v \geq \sqrt{\frac{h}{4\pi m}} \]

7. **Calculate Δp**:
Now we can find the uncertainty in momentum (Δp):
\[ \Delta p = m \cdot \Delta v \]
Substituting Δv from the previous step:
\[ \Delta p \geq m \cdot \sqrt{\frac{h}{4\pi m}} \]
This simplifies to:
\[ \Delta p \geq \sqrt{m \cdot \frac{h}{4\pi}} \]

### Final Answer:
Thus, the uncertainty in momentum is:
\[ \Delta p = \sqrt{\frac{hm}{4\pi}} \]