If position and momentum of an electron are determined at a time then value of `DeltaP` is obtained `3DeltaX`, now the uncertainity
in velocity of electron will be `:-`

Options

A. `(1)/(2m)sqrt((h)/(3pi))`
B. `(1)/(2m)sqrt((3h)/(pi))`
C. `(1)/(4m)sqrt((3h)/(pi))`
D. `(1)/(6m)sqrt((h)/(pi))`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between momentum and velocity
The momentum \( P \) of an electron is given by the equation:
\[ P = mv \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

### Step 2: Express the uncertainty in momentum
The uncertainty in momentum \( \Delta P \) can be expressed in terms of the uncertainty in velocity \( \Delta v \):
\[ \Delta P = m \Delta v \]

### Step 3: Relate the uncertainties
According to the problem, we have:
\[ \Delta P = 3 \Delta x \]
where \( \Delta x \) is the uncertainty in position.

### Step 4: Set the equations equal
From the above relationships, we can set the equations for \( \Delta P \) equal to each other:
\[ m \Delta v = 3 \Delta x \]

### Step 5: Solve for \( \Delta v \)
Rearranging the equation gives:
\[ \Delta v = \frac{3 \Delta x}{m} \]

### Step 6: Apply Heisenberg's uncertainty principle
According to Heisenberg's uncertainty principle, we have:
\[ \Delta P \cdot \Delta x \geq \frac{h}{4\pi} \]
Substituting \( \Delta P = 3 \Delta x \) into the uncertainty principle gives:
\[ 3 \Delta x \cdot \Delta x \geq \frac{h}{4\pi} \]
This simplifies to:
\[ 3 (\Delta x)^2 \geq \frac{h}{4\pi} \]

### Step 7: Solve for \( \Delta x \)
From this inequality, we can express \( \Delta x \):
\[ (\Delta x)^2 \geq \frac{h}{12\pi} \]
Taking the square root gives:
\[ \Delta x \geq \sqrt{\frac{h}{12\pi}} \]

### Step 8: Substitute \( \Delta x \) back to find \( \Delta v \)
Now substituting \( \Delta x \) back into the equation for \( \Delta v \):
\[ \Delta v = \frac{3 \Delta x}{m} \geq \frac{3}{m} \sqrt{\frac{h}{12\pi}} \]

### Final Answer
Thus, the uncertainty in the velocity of the electron is:
\[ \Delta v \geq \frac{3 \sqrt{h}}{m \sqrt{12\pi}} \]