An electron is allowed to move freely in a closed cubic box of length 10 cm. The minimum uncertainty in its velocity will be observed as:
Correct answer: B. 5.8×10⁻⁴ m/s
- A. 4×10⁻³ m/s
- B. 5.8×10⁻⁴ m/s
- C. 4×10⁻⁵ m/s
- D. 4×10⁻⁶ m/s
Explanation
According to Heisenberg's uncertainty principle, Δx·Δp ≥ h/4π. The uncertainty in position Δx is the length of the box (0.1 m). Solving for the uncertainty in velocity (Δv = Δp/m). gives Δv ≥ h / (4πmΔx), which calculates to approximately 5.8 × 10⁻⁴ m/s.
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