Using Heisenberg's uncertainty principle, calculate the uncertainty in velocity of an electron if uncertainty in its position is `10^(-11)m` Given, `h =6.6 xx 10^(-14)kg m^2s^(-1), m=9.1 xx 10^(-31)kg`

Answer

To solve the problem using Heisenberg's uncertainty principle, we will follow these steps:

### Step 1: Understand Heisenberg's Uncertainty Principle
The 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, which is given by:

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

Where:
- \( \Delta x \) = uncertainty in position
- \( \Delta p \) = uncertainty in momentum
- \( h \) = Planck's constant

### Step 2: Relate Momentum to Velocity
Momentum (p) is defined as the product of mass (m) and velocity (v):

\[ p = m \cdot v \]

Thus, the uncertainty in momentum can be expressed as:

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

Where \( \Delta v \) is the uncertainty in velocity.

### Step 3: Substitute into the Uncertainty Principle
Substituting the expression for Δp into the uncertainty principle gives:

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

### Step 4: Solve for Uncertainty in Velocity
Rearranging the equation to solve for Δv:

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

### Step 5: Substitute the Given Values
Now, we can substitute the values provided in the question:

- \( h = 6.6 \times 10^{-34} \, \text{kg m}^2/\text{s} \)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- \( \Delta x = 10^{-11} \, \text{m} \)

Substituting these values into the equation:

\[ \Delta v \geq \frac{6.6 \times 10^{-34}}{4\pi \cdot 9.1 \times 10^{-31} \cdot 10^{-11}} \]

### Step 6: Calculate the Value
Now, calculate the denominator:

1. Calculate \( 4\pi \):
\[ 4\pi \approx 12.566 \]

2. Calculate \( 4\pi \cdot 9.1 \times 10^{-31} \cdot 10^{-11} \):
\[ 4\pi \cdot 9.1 \times 10^{-31} \cdot 10^{-11} \approx 12.566 \cdot 9.1 \times 10^{-42} \approx 1.144 \times 10^{-41} \]

3. Now substitute back to find Δv:
\[ \Delta v \geq \frac{6.6 \times 10^{-34}}{1.144 \times 10^{-41}} \approx 5.77 \times 10^{6} \, \text{m/s} \]

### Final Answer
Thus, the uncertainty in the velocity of the electron is:

\[ \Delta v \approx 5.77 \times 10^{6} \, \text{m/s} \]

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