The uncertainty in the velocity of a particle of mass `6.626 xx 10^(-31) kg` is `1 xx 10^6 ms^(-1)`. What is the uncertainty in its position (in nm) ? `(h = 6.626 xx 10^(-34) Js)`

Options

A. `(1/(2pi))`
B. `((2.5)/(pi))`
C. `(4/(pi))`
D. `(1/(4pi))`

Explanation

To find the uncertainty in the position of a particle using the Heisenberg Uncertainty Principle, we can follow these steps:

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

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

Given:
- Mass \( m = 6.626 \times 10^{-31} \, \text{kg} \)
- Uncertainty in velocity \( \Delta v = 1 \times 10^{6} \, \text{m/s} \)

Now we can calculate \( \Delta p \):
\[ \Delta p = (6.626 \times 10^{-31} \, \text{kg}) \cdot (1 \times 10^{6} \, \text{m/s}) = 6.626 \times 10^{-25} \, \text{kg m/s} \]

### Step 3: Substitute into the uncertainty principle
Now we substitute \( \Delta p \) into the uncertainty principle equation:
\[ \Delta x \cdot (6.626 \times 10^{-25}) \geq \frac{6.626 \times 10^{-34}}{4\pi} \]

### Step 4: Calculate \( \frac{h}{4\pi} \)
Calculating \( \frac{h}{4\pi} \):
\[ \frac{6.626 \times 10^{-34}}{4\pi} \approx \frac{6.626 \times 10^{-34}}{12.5664} \approx 5.283 \times 10^{-35} \, \text{Js} \]

### Step 5: Solve for \( \Delta x \)
Now we can solve for \( \Delta x \):
\[ \Delta x \geq \frac{5.283 \times 10^{-35}}{6.626 \times 10^{-25}} \approx 7.97 \times 10^{-11} \, \text{m} \]

### Step 6: Convert to nanometers
To convert meters to nanometers (1 nm = \( 1 \times 10^{-9} \, \text{m} \)):
\[ \Delta x \geq 7.97 \times 10^{-11} \, \text{m} = 0.0797 \, \text{nm} \]

### Final Answer
The uncertainty in the position of the particle is approximately:
\[ \Delta x \geq 0.0797 \, \text{nm} \]

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