The uncertainty in velocity of an electron present in the nucleous of diameter `10 ^(-15)` , 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 solve the problem of finding the uncertainty in the velocity of an electron present in the nucleus of diameter \(10^{-15}\) m, we will use the Heisenberg Uncertainty Principle. The principle states that:

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

Where:
- \(\Delta x\) is the uncertainty in position,
- \(\Delta p\) is the uncertainty in momentum,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)).

### Step-by-Step Solution:

1. **Determine \(\Delta x\)**:
The uncertainty in position \(\Delta x\) can be approximated as the diameter of the nucleus:
\[ \Delta x = 10^{-15} \, \text{m} \]

2. **Relate \(\Delta p\) to \(\Delta v\)**:
The uncertainty in momentum \(\Delta p\) can be expressed in terms of mass \(m\) and uncertainty in velocity \(\Delta v\):
\[ \Delta p = m \cdot \Delta v \]
Where the mass of the electron \(m\) is approximately \(9.1 \times 10^{-31} \, \text{kg}\).

3. **Substitute into the Uncertainty Principle**:
Substitute \(\Delta x\) and \(\Delta p\) into the Heisenberg Uncertainty Principle:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]
This can be rearranged to solve for \(\Delta v\):
\[ \Delta v \geq \frac{h}{4\pi m \Delta x} \]

4. **Plug in the Values**:
Now, substitute the known values into the equation:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4\pi (9.1 \times 10^{-31}) (10^{-15})} \]

5. **Calculate**:
Calculate the right-hand side:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15}} \]
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{1.136 \times 10^{-44}} \approx 5.83 \times 10^{10} \, \text{m/s} \]

6. **Final Result**:
Therefore, the uncertainty in the velocity of the electron is approximately:
\[ \Delta v \approx 5.83 \times 10^{10} \, \text{m/s} \approx 10^{11} \, \text{m/s} \]

### Conclusion:
The uncertainty in the velocity of an electron present in the nucleus of diameter \(10^{-15}\) m is approximately \(10^{11} \, \text{m/s}\).