Given `m_(e)=9.11 xx 10^(-31)kg` and `h = 6.626 xx 10^(-34)Js`, the uncertainty involved in the measurement of velocity within a distance of `0.1Å` is

Options

A. `5.79 xx 10^(8) ms^(-1)`
B. `5.79 xx 10^(5) ms^(-1)`
C. `5.79 xx 10^(6) ms^(-1)`
D. `5.79 xx 10^(7) ms^(-1)`

Explanation

To find the uncertainty involved in the measurement of velocity within a distance of \(0.1 \, \text{Å}\), we can use the Heisenberg Uncertainty Principle, which states:

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

Where:
- \(\Delta x\) is the uncertainty in position,
- \(\Delta p\) is the uncertainty in momentum,
- \(h\) is the Planck constant.

1. **Convert the distance from angstroms to meters**:
\[ 0.1 \, \text{Å} = 0.1 \times 10^{-10} \, \text{m} = 1.0 \times 10^{-11} \, \text{m} \]

2. **Substitute \(\Delta x\) into the uncertainty principle**:
\[ \Delta p = \frac{h}{4\pi \Delta x} \]
Substituting the values:
\[ h = 6.626 \times 10^{-34} \, \text{Js}, \quad \Delta x = 1.0 \times 10^{-11} \, \text{m} \]
\[ \Delta p = \frac{6.626 \times 10^{-34}}{4 \cdot \pi \cdot 1.0 \times 10^{-11}} \]

3. **Calculate the value of \(\Delta p\)**:
\[ \Delta p = \frac{6.626 \times 10^{-34}}{4 \cdot 3.14 \cdot 1.0 \times 10^{-11}} \]
\[ \Delta p = \frac{6.626 \times 10^{-34}}{1.256 \times 10^{-10}} \]
\[ \Delta p \approx 5.28 \times 10^{-24} \, \text{kg m/s} \]

4. **Relate the uncertainty in momentum to the uncertainty in velocity**:
\[ \Delta p = m \cdot \Delta v \]
Where \(m\) is the mass of the electron, \(m_e = 9.11 \times 10^{-31} \, \text{kg}\).

5. **Solve for \(\Delta v\)**:
\[ \Delta v = \frac{\Delta p}{m} \]
\[ \Delta v = \frac{5.28 \times 10^{-24}}{9.11 \times 10^{-31}} \]
\[ \Delta v \approx 5.79 \times 10^{6} \, \text{m/s} \]

Thus, the uncertainty in the measurement of velocity within a distance of \(0.1 \, \text{Å}\) is approximately \(5.79 \times 10^{6} \, \text{m/s}\).