If an electron is , to be located within `10 pm` what will be the uncertainty in its velocity ?

Explanation

To find the uncertainty in the velocity of an electron when it is located within 10 picometers (pm), we can use Heisenberg's Uncertainty Principle. Let's solve it step by step.

### Step 1: Understand Heisenberg's Uncertainty Principle
The principle states that the product of the uncertainties in position (Δx) and momentum (Δp) is greater than or equal to a constant:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
where:
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \( \Delta x \) is the uncertainty in position,
- \( \Delta p \) is the uncertainty in momentum.

### Step 2: Convert the Uncertainty in Position
Given that the uncertainty in position is \(10 \, \text{pm}\):
\[ \Delta x = 10 \, \text{pm} = 10 \times 10^{-12} \, \text{m} = 10^{-11} \, \text{m} \]

### Step 3: Relate Uncertainty in 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 4: Substitute into the Uncertainty Principle
Substituting \( \Delta p \) into the uncertainty principle gives:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]

### Step 5: Rearrange to Solve for Uncertainty in Velocity
Rearranging the equation to solve for \( \Delta v \):
\[ \Delta v \geq \frac{h}{4\pi m \Delta x} \]

### Step 6: Plug in the Values
Now we can plug in the values:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- \( \Delta x = 10^{-11} \, \text{m} \)

Calculating:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4 \cdot \pi \cdot (9.1 \times 10^{-31}) \cdot (10^{-11})} \]

### Step 7: Calculate the Numerical Value
Calculating the denominator:
\[ 4 \cdot \pi \cdot (9.1 \times 10^{-31}) \cdot (10^{-11}) \approx 1.141 \times 10^{-40} \]
Now substituting this back into the equation:
\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{1.141 \times 10^{-40}} \approx 5.8 \times 10^{6} \, \text{m/s} \]

### Final Answer
The uncertainty in the velocity of the electron is approximately:
\[ \Delta v \geq 5.8 \times 10^{6} \, \text{m/s} \]

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