Question

# Calculate the uncertainty in the position (Δx) of an electron if Δv is 0.1%. Take the velocity of electron =2.2×106 ms−1 and mass of electron as 9.108×10−31kg.

A

262.48 × 10-10 m

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B

9.1 × 10-9 m

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C

458.76 × 10-12 m

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D

959.67 × 10-10 m

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Solution

## The correct option is A 262.48 × 10-10 m From Heisenberg Uncertainty Principle, we know Δp × Δx = h4π ⇒ Δv × Δx = h4mπ Now, Given , Δv=0.1% of the velocity of the electron =0.1100 ×2.2 × 106 =2.2× 103ms−1 Δx×mΔv=h4π or Δx=6.63×10−34Js4×3.14×9.108×10−31kg×2.2×103ms−1 =0.02624765×10−6m =262.4765×10−10m Since Δx is much longer than the atomic diameter (≈ 10−10m), the uncertainty principle is applicable in this case.

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