Heisenberg Uncertainty Principle
If the uncertainty in the position of an electron is 0.1 nm, the minimum uncertainty in its momentum is approximately (h = 6.63 × 10⁻³⁴ J s):
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About This Question
- Subject
- chemistry
- Chapter
- atomic structure
- Topic
- heisenberg uncertainty principle
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
5.27 × 10⁻²⁵ kg m s⁻¹
According to Heisenberg's uncertainty principle, Δx · Δp ≥ h/4π. Given Δx = 0.1 nm = 1 × 10⁻¹⁰ m, the minimum uncertainty in momentum is Δp = h/(4πΔx) = 6.63 × 10⁻³⁴ / (4 × 3.14159 × 1 × 10⁻¹⁰) = 6.63 × 10⁻³⁴ / 12.566 × 10⁻¹⁰ = 5.27 × 10⁻²⁵ kg m s⁻¹. This principle highlights that we cannot simultaneously determine both the exact position and momentum of a subatomic particle — greater precision in one quantity leads to greater uncertainty in the other.
This medium difficulty chemistry question is from the chapter atomic structure, covering the topic of heisenberg uncertainty principle. It appeared in the 2025 exam.
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