Question

If the position of the electron was measured with an accuracy of \( \pm 0.002 \) nm, the uncertainty in the momentum of it would be (in kg ms\(^{-1}\)) (\( h = 6.626 \times 10^{-34} \) Js):

• A. \( 2.637 \times 10^{-23} \)
• B. \( 2.637 \times 10^{-24} \)
• C. \( 8.283 \times 10^{-23} \)
• D. \( 8.283 \times 10^{-24} \)

Hint:

Heisenberg's uncertainty principle states that the more precisely we know a particle's position, the less precisely we can know its momentum.

Answer 1

The Correct Option is A

Step 1: Use Heisenberg's Uncertainty Principle \[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \] where: 
- \( \Delta x = 0.002 \) nm \( = 2 \times 10^{-12} \) m, - \( h = 6.626 \times 10^{-34} \) Js. 
Step 2: Compute \( \Delta p \) \[ \Delta p = \frac{h}{4\pi \Delta x} \] \[ = \frac{6.626 \times 10^{-34}}{4\pi \times 2 \times 10^{-12}} \] \[ = 2.637 \times 10^{-23} \text{ kg ms}^{-1} \]