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ExamsJEE MainPhysics

The uncertainty in the momentum of an electron is 1.0 * 10⁻⁵ kg*m/s. Using the Heisenberg uncertainty principle, what is the minimum uncertainty in its position? (h = 6.626 * 10⁻³⁴ J*s)

  1. 1.05 * 10⁻²⁸ m
  2. 1.05 * 10⁻²⁶ m
  3. 5.27 * 10⁻³⁰ m
  4. 5.25 * 10⁻²⁸ m

Correct answer: 5.27 * 10⁻³⁰ m

Solution

Minimum uncertainty: deltaₓ = h/(4*pi*deltaₚ) = 6.626*10⁻³⁴ / (4 * 3.1416 * 1.0*10⁻⁵) = 6.626*10⁻³⁴ / (1.2566*10⁻⁴) = 5.27*10⁻³⁰ m.

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