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 find the uncertainty in the component of the velocity (Δv) for an electron given its positional uncertainty (Δx), we can use the Heisenberg Uncertainty Principle, which states:

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

Where:
- Δx is the uncertainty in position,
- Δp is the uncertainty in momentum,
- h is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)).

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

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

Substituting this into the uncertainty principle gives:

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

From this, we can rearrange to find Δv:

\[ \Delta v \geq \frac{h}{4\pi m \Delta x} \]

Now, we can substitute the known values:
- \(h = 6.626 \times 10^{-34} \, \text{Js}\)
- \(m = 9.11 \times 10^{-31} \, \text{kg}\)
- \(\Delta x = 1.0 \times 10^{-10} \, \text{m}\)

Now we can calculate Δv:

1. Substitute the values into the equation:

\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4 \cdot \pi \cdot 9.11 \times 10^{-31} \cdot 1.0 \times 10^{-10}} \]

2. Calculate \(4 \cdot \pi\):

\[ 4 \cdot \pi \approx 12.566 \]

3. Now calculate the denominator:

\[ 4 \cdot \pi \cdot 9.11 \times 10^{-31} \cdot 1.0 \times 10^{-10} \approx 12.566 \cdot 9.11 \times 10^{-41} \approx 1.144 \times 10^{-40} \]

4. Now substitute this back into the equation for Δv:

\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{1.144 \times 10^{-40}} \approx 5.79 \times 10^{6} \, \text{m/s} \]

Thus, the uncertainty in the component of the velocity is approximately:

\[ \Delta v \approx 5.79 \times 10^{6} \, \text{m/s} \]