The uncertainty in velocity of an electron present in the nucleous of diameter `10 ^(-15)` m hypothetically should be approximately

Options

A. `10^(-11)`m/s
B. `10^8` m/s
C. `10^11` m/s
D. `10Å//s`

Answer

To find the uncertainty in the velocity of an electron present in the nucleus of diameter \(10^{-15}\) m, we can use the Heisenberg Uncertainty Principle, which is mathematically expressed as:

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

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

### Step 1: Identify the values
- The diameter of the nucleus is given as \(10^{-15}\) m. Therefore, we can assume the uncertainty in position (\(\Delta x\)) is approximately equal to this diameter:
\[ \Delta x = 10^{-15} \, \text{m} \]
- The mass of the electron (\(m\)) is approximately \(9.1 \times 10^{-31} \, \text{kg}\).

### Step 2: Rearranging the formula
We can rearrange the Heisenberg Uncertainty Principle to solve for \(\Delta v\):
\[ \Delta v \geq \frac{h}{4\pi m \Delta x} \]

### Step 3: Substitute the known values
Now, substituting the known values into the equation:
\[ \Delta v \geq \frac{6.63 \times 10^{-34}}{4 \cdot 3.14 \cdot (9.1 \times 10^{-31}) \cdot (10^{-15})} \]

### Step 4: Calculate the denominator
Calculating the denominator:
\[ 4 \cdot 3.14 \approx 12.56 \]
\[ 12.56 \cdot (9.1 \times 10^{-31}) \cdot (10^{-15}) \approx 1.143 \times 10^{-45} \]

### Step 5: Calculate \(\Delta v\)
Now substituting back into the equation:
\[ \Delta v \geq \frac{6.63 \times 10^{-34}}{1.143 \times 10^{-45}} \approx 5.79 \times 10^{11} \, \text{m/s} \]

### Conclusion
Thus, the uncertainty in the velocity of the electron is approximately:
\[ \Delta v \approx 0.58 \times 10^{11} \, \text{m/s} \]

### Final Answer
The uncertainty in velocity of the electron is approximately \(5.8 \times 10^{10} \, \text{m/s}\).