If an electron is travelling at 200 m/s within 1 m/s uncertainty, whtat is the theoretical uncertainty in its position in mum (micrometer)?

Options

A. 14.5
B. 29
C. 58
D. 114

Answer

To solve the problem of determining the theoretical uncertainty in the position of an electron traveling at a velocity of 200 m/s with an uncertainty of 1 m/s, we will use Heisenberg's uncertainty principle. Here’s a step-by-step solution:

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

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

### Step 3: Rearranging the Equation
From the uncertainty principle, we can rearrange the equation to solve for the uncertainty in position (Δx):
\[ \Delta x \geq \frac{h}{4\pi m \Delta v} \]

### Step 4: Substitute Known Values
We know the following values:
- Planck's constant \(h = 6.626 \times 10^{-34} \, \text{J s}\)
- Mass of the electron \(m = 9.1 \times 10^{-31} \, \text{kg}\)
- Uncertainty in velocity \(\Delta v = 1 \, \text{m/s}\)

Now, substituting these values into the equation:
\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{4 \cdot \pi \cdot (9.1 \times 10^{-31}) \cdot (1)} \]

### Step 5: Calculate the Denominator
Calculating the denominator:
\[ 4 \cdot \pi \cdot (9.1 \times 10^{-31}) \approx 4 \cdot 3.14 \cdot 9.1 \times 10^{-31} \approx 1.12 \times 10^{-29} \]

### Step 6: Calculate Δx
Now substituting this value back into the equation for Δx:
\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{1.12 \times 10^{-29}} \approx 5.91 \times 10^{-6} \, \text{m} \]

### Step 7: Convert to Micrometers
To convert meters to micrometers (1 micrometer = \(10^{-6}\) meters):
\[ \Delta x \approx 5.91 \, \mu m \]

### Step 8: Final Result
Thus, the theoretical uncertainty in the position of the electron is approximately:
\[ \Delta x \approx 5.91 \, \mu m \]