What will be the uncertainty in velocity of an electrons when the uncertainty in its positions is 1000 Å?

Options

A. `5.79 xx 10^2 ms^(-1)`
B. `5.79 xx 10^(8) ms^(-1)`
C. `5.79 xx 10^4 ms^(-1)`
D. `5.79 xx 10^(-10)ms^(-1)`

Answer

To find the uncertainty in the velocity of an electron when the uncertainty in its position is given, we will use the Heisenberg Uncertainty Principle. The principle states that the product of the uncertainties in position and momentum of a particle cannot be smaller than a certain value.

### Step-by-Step Solution:

1. **Identify the given values:**
- Uncertainty in position (Δx) = 1000 Å = 1000 × 10^(-10) m = 10^(-7) m.
- Mass of the electron (m) = 9.1 × 10^(-31) kg.

2. **Write the Heisenberg Uncertainty Principle formula:**
The Heisenberg Uncertainty Principle can be expressed as:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
where Δp is the uncertainty in momentum and h is Planck's constant (h ≈ 6.626 × 10^(-34) Js).

3. **Relate momentum to velocity:**
The momentum (p) of an electron is given by:
\[ p = mv \]
Therefore, the uncertainty in momentum (Δp) can be expressed as:
\[ \Delta p = m \cdot \Delta v \]
where Δv is the uncertainty in velocity.

4. **Substitute Δp in the Heisenberg formula:**
Substitute Δp into the Heisenberg Uncertainty Principle:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

5. **Rearrange the equation to solve for Δv:**
\[ \Delta v \geq \frac{h}{4\pi m \Delta x} \]

6. **Substitute the known values:**
- Planck's constant (h) = 6.626 × 10^(-34) Js
- Mass of the electron (m) = 9.1 × 10^(-31) kg
- Uncertainty in position (Δx) = 10^(-7) m

Now substitute these values into the equation:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4 \pi (9.1 \times 10^{-31}) (10^{-7})} \]

7. **Calculate Δv:**
First, calculate the denominator:
\[ 4 \pi (9.1 \times 10^{-31}) (10^{-7}) \approx 1.141 \times 10^{-37} \]
Now calculate Δv:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{1.141 \times 10^{-37}} \approx 5.79 \times 10^{3} \text{ m/s} \]

### Final Answer:
The uncertainty in the velocity of the electron is approximately:
\[ \Delta v \approx 5.79 \times 10^{3} \text{ m/s} \]