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

Options

A. `5.79 xx10^(6)` m sec^(-1)`
B. `5.79 xx10^(7) m sec^(-1)`
C. `5.79 xx10^(8) m sec^(-1)`
D. `5.79 xx10^(5) m sec^(-1)`

Answer

To solve the problem of finding the uncertainty involved in the measurement of the velocity of an electron within a distance of \(0.1 \, \text{Å}\), we will use the Heisenberg Uncertainty Principle. Here’s a step-by-step solution:

### Step 1: Understand the Heisenberg Uncertainty Principle
The Heisenberg Uncertainty Principle states that the product of the uncertainties in position (\(\Delta x\)) and momentum (\(\Delta p\)) is greater than or equal to a constant:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
where \(h\) is Planck's constant.

### Step 2: Relate Momentum to Velocity
Momentum (\(p\)) is defined as the product of mass (\(m\)) and velocity (\(v\)):
\[ p = mv \]
Thus, the uncertainty in momentum can be expressed as:
\[ \Delta p = m \cdot \Delta v \]
where \(\Delta v\) is the uncertainty in velocity.

### Step 3: Substitute into the Uncertainty Principle
Substituting \(\Delta p\) into the Heisenberg Uncertainty Principle gives:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

### Step 4: Solve for \(\Delta v\)
Rearranging the equation to solve for \(\Delta v\):
\[ \Delta v \geq \frac{h}{4\pi m \Delta x} \]

### Step 5: Insert Known Values
Now, we will substitute the known values:
- Planck's constant, \(h = 6.626 \times 10^{-34} \, \text{Js}\)
- Mass of the electron, \(m = 9.11 \times 10^{-31} \, \text{kg}\)
- Uncertainty in position, \(\Delta x = 0.1 \, \text{Å} = 0.1 \times 10^{-10} \, \text{m}\)

Substituting these values into the equation:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4 \cdot 3.14 \cdot (9.11 \times 10^{-31}) \cdot (0.1 \times 10^{-10})} \]

### Step 6: Calculate the Right Side
Calculating the denominator:
\[ 4 \cdot 3.14 \cdot (9.11 \times 10^{-31}) \cdot (0.1 \times 10^{-10}) \approx 1.151 \times 10^{-41} \]
Now, calculate \(\Delta v\):
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{1.151 \times 10^{-41}} \approx 5.76 \times 10^{7} \, \text{m/s} \]

### Step 7: Final Result
Thus, the uncertainty in the measurement of the velocity of the electron is:
\[ \Delta v \geq 5.76 \times 10^{6} \, \text{m/s} \]

### Conclusion
The uncertainty involved in the measurement of the velocity of an electron within a distance of \(0.1 \, \text{Å}\) is approximately \(5.76 \times 10^{6} \, \text{m/s}\).