Question

Uncertainty principle is valid for:

• A. Proton
• B. Methane
• C. Both (A) and (B)
• D. \( 1 \, \mu \text{m} \) sized platinum particles
• E. \( 1 \, \mu \text{m} \) sized NaCl particles

Hint:

The uncertainty principle primarily applies to subatomic particles, where quantum effects are significant. For macroscopic objects like molecules or large particles, the uncertainty is negligible.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Heisenberg's Uncertainty Principle declares that it is fundamentally impossible to simultaneously determine both the exact position and momentum of a particle.
This principle is a core component of quantum mechanics and is significant primarily for microscopic objects.
Step 2: Key Formula or Approach:
The mathematical expression for Heisenberg's uncertainty principle is \( \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \).
The mass of the object dictates how observable this uncertainty is.
Step 3: Detailed Explanation:
While the principle theoretically applies to all matter, its practical significance is inversely proportional to the mass of the object.
For macroscopic objects, the mass is so large that the uncertainty in position and velocity becomes infinitesimally small and completely unobservable.
A proton is a subatomic particle, meaning it has an extremely small mass and exhibits significant quantum uncertainty.
A methane molecule is also a microscopic entity on the atomic scale, making the uncertainty principle valid and significant for it.
Particles of size 1\(\mu\)m (like the platinum or NaCl particles) are considered macroscopic in the context of quantum mechanics, and their quantum uncertainties are negligible.
Therefore, the principle holds true measurable validity for both protons and methane.
Step 4: Final Answer:
The uncertainty principle is valid for both the proton and the methane molecule.