An electron is allowed to move freely in a closed cubic box of length of side 10 cm. The uncertainty in its velocity will be :

Options

A. (a) `3.35xx10^(-4)" m sec"^(-1)`
B. (b) `5.8xx10^(-4)" m sec"^(-1)`
C. (c) `4xx10^(-5)" m sec "^(-1)`
D. (d) `4xx10^(-6)" m sec "^(-1)`

Answer

To find the uncertainty in the velocity of an electron moving freely in a closed cubic box of side length 10 cm, we can use Heisenberg's Uncertainty Principle.

### Step-by-Step Solution:

1. **Convert the Length of the Cube to Meters:**
The side length of the cube is given as 10 cm. We need to convert this to meters for standard SI units.
\[ \text{Length of the cube} = 10 \, \text{cm} = 0.1 \, \text{m} \]

2. **Identify the Uncertainty in Position (Δx):**
Since the electron can move freely within the cube, the uncertainty in its position (Δx) is equal to the length of the side of the cube.
\[ \Delta x = 0.1 \, \text{m} \]

3. **Apply Heisenberg's Uncertainty Principle:**
According to Heisenberg's Uncertainty Principle:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
where Δp is the uncertainty in momentum and h is Planck's constant.

4. **Express Momentum in Terms of Velocity:**
Momentum (p) is given by the product of mass (m) and velocity (v):
\[ p = m \cdot v \]
Therefore, the uncertainty in momentum (Δp) can be expressed as:
\[ \Delta p = m \cdot \Delta v \]

5. **Substitute Δp into the Uncertainty Principle:**
Substituting Δp into the uncertainty principle gives:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

6. **Rearranging for Δv:**
Rearranging the equation to solve for the uncertainty in velocity (Δv):
\[ \Delta v \geq \frac{h}{4\pi \Delta x m} \]

7. **Insert Known Values:**
- Planck's constant (h) = \(6.626 \times 10^{-34} \, \text{Js}\)
- Mass of the electron (m) = \(9.109 \times 10^{-31} \, \text{kg}\)
- Δx = 0.1 m

Substituting these values into the equation:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4 \cdot 3.14 \cdot 0.1 \cdot 9.109 \times 10^{-31}} \]

8. **Calculate Δv:**
Performing the calculations:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{1.151 \times 10^{-31}} \approx 5.76 \times 10^{-4} \, \text{m/s} \]

9. **Final Result:**
The uncertainty in the velocity of the electron is approximately:
\[ \Delta v \approx 5.76 \times 10^{-4} \, \text{m/s} \]