The uncertainty involved in the measurement of velocity of electron with a distance of `0.1Å` is

Options

A. `5.79xx10^(8)" m/s"`
B. `5.79xx10^(5)" m/s"`
C. `5.79xx10^(6)" m/s"`
D. `5.79xx10^(7)" m/s"`

Explanation

To solve the problem of finding the uncertainty involved in the measurement of the velocity of an electron with a distance of `0.1 Å`, we will apply the Heisenberg Uncertainty Principle. Here are the steps to arrive at the solution:

### Step 1: Understand the Heisenberg Uncertainty Principle
The Heisenberg Uncertainty Principle states that the product of the uncertainties in position (Δx) and momentum (Δp) of a particle is given by the formula:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
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 2: Convert the given distance to meters
The distance given is `0.1 Å` (angstrom). We need to convert this to meters:
\[ 0.1 \, \text{Å} = 0.1 \times 10^{-10} \, \text{m} = 1 \times 10^{-11} \, \text{m} \]
Thus, \( \Delta x = 1 \times 10^{-11} \, \text{m} \).

### Step 3: Calculate the uncertainty in momentum
The uncertainty in momentum (Δp) can be expressed as:
\[ \Delta p = m \cdot \Delta v \]
where:
- \( m \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \)),
- \( \Delta v \) is the uncertainty in velocity.

### Step 4: Substitute values into the Heisenberg equation
Substituting \( \Delta x \) and \( \Delta p \) into the Heisenberg Uncertainty Principle equation:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]
\[ 1 \times 10^{-11} \cdot (9.1 \times 10^{-31} \cdot \Delta v) \geq \frac{6.626 \times 10^{-34}}{4\pi} \]

### Step 5: Calculate the right-hand side
Calculating \( \frac{h}{4\pi} \):
\[ \frac{6.626 \times 10^{-34}}{4 \cdot 3.14} \approx \frac{6.626 \times 10^{-34}}{12.56} \approx 5.28 \times 10^{-35} \]

### Step 6: Rearrange to find Δv
Now, we can rearrange the equation to solve for \( \Delta v \):
\[ 1 \times 10^{-11} \cdot (9.1 \times 10^{-31} \cdot \Delta v) \geq 5.28 \times 10^{-35} \]
\[ \Delta v \geq \frac{5.28 \times 10^{-35}}{9.1 \times 10^{-31} \cdot 1 \times 10^{-11}} \]
\[ \Delta v \geq \frac{5.28 \times 10^{-35}}{9.1 \times 10^{-42}} \]
\[ \Delta v \geq 5.79 \times 10^{6} \, \text{m/s} \]

### Conclusion
The uncertainty in the measurement of the velocity of the electron is:
\[ \Delta v \geq 5.79 \times 10^{6} \, \text{m/s} \]