For an electron whose positional uncertainly is `1.0xx10^(-10)m`, the uncertainty in the component of the velocity in `ms^(-1)`will be

Options

A. `5.8xx10^(5)`
B. `10^(9)`
C. `10^(2)`
D. `10^(15)`

Answer

To solve the problem of finding the uncertainty in the component of the velocity of an electron given its positional uncertainty, we will use the Heisenberg Uncertainty Principle. Here's the step-by-step 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 at least on the order of Planck's constant divided by 4π:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]

### Step 2: Define the variables
- Given positional uncertainty (Δx) = \(1.0 \times 10^{-10} \, m\)
- Planck's constant (h) = \(6.626 \times 10^{-34} \, J \cdot s\)
- Mass of the electron (m) = \(9.1 \times 10^{-31} \, kg\)

### Step 3: Relate momentum and velocity
Momentum (p) is related to velocity (v) by the equation:
\[ p = mv \]
Thus, the uncertainty in momentum (Δp) can be expressed as:
\[ \Delta p = m \Delta v \]

### Step 4: Substitute Δp into the uncertainty principle
Substituting Δp into the Heisenberg Uncertainty Principle gives:
\[ \Delta x \cdot (m \Delta v) = \frac{h}{4\pi} \]

### Step 5: Solve for Δv
Rearranging the equation to solve for the uncertainty in velocity (Δv):
\[ \Delta v = \frac{h}{4\pi m \Delta x} \]

### Step 6: Plug in the values
Now, substituting the known values into the equation:
\[ \Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times (9.1 \times 10^{-31}) \times (1.0 \times 10^{-10})} \]

### Step 7: Calculate Δv
Calculating the denominator:
\[ 4 \times 3.14 \times 9.1 \times 10^{-31} \times 1.0 \times 10^{-10} \approx 1.134 \times 10^{-40} \]
Now calculating Δv:
\[ \Delta v \approx \frac{6.626 \times 10^{-34}}{1.134 \times 10^{-40}} \approx 5.84 \times 10^{6} \, m/s \]

### Step 8: Final Result
Thus, the uncertainty in the component of the velocity of the electron is approximately:
\[ \Delta v \approx 5.8 \times 10^{5} \, m/s \]

### Conclusion
The answer is:
\[ \Delta v \approx 5.8 \times 10^{5} \, m/s \]