What is the uncertainty in velocity of an electron if the uncertainty in its position is `10^(-10) m`? Mass of the electron is `9.1 xx 10^(-31) kg and h = 6.6 xx 10^(-34) m^(2) s^(-1)`?

Explanation

To find the uncertainty in the velocity of an electron given the uncertainty in its position, we can 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.

Since momentum (\(p\)) is defined as the product of mass (\(m\)) and velocity (\(v\)), we can express the uncertainty in momentum as:

\[ \Delta p = m \cdot \Delta v \]

where \(\Delta v\) is the uncertainty in velocity.

### Step-by-Step Solution:

1. **Identify the given values**:
- Uncertainty in position, \(\Delta x = 10^{-10} \, \text{m}\)
- Mass of the electron, \(m = 9.1 \times 10^{-31} \, \text{kg}\)
- Planck's constant, \(h = 6.626 \times 10^{-34} \, \text{m}^2 \text{s}^{-1}\)

2. **Set up the Heisenberg Uncertainty Principle**:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

3. **Rearrange the equation to solve for \(\Delta v\)**:
\[ \Delta v \geq \frac{h}{4\pi \Delta x \cdot m} \]

4. **Substitute the known values into the equation**:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4\pi \cdot (10^{-10}) \cdot (9.1 \times 10^{-31})} \]

5. **Calculate the denominator**:
- Calculate \(4\pi \approx 12.566\)
- Therefore, the denominator becomes:
\[ 12.566 \cdot (10^{-10}) \cdot (9.1 \times 10^{-31}) \approx 1.143 \times 10^{-40} \]

6. **Calculate \(\Delta v\)**:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{1.143 \times 10^{-40}} \approx 5.79 \times 10^{6} \, \text{m/s} \]

### Final Result:
The uncertainty in the velocity of the electron is approximately:
\[ \Delta v \approx 5.79 \times 10^{6} \, \text{m/s} \]