Particle as a wave

Task number: 4404

Assume that the uncertainty of a particle’s location is equal to its de Broglie wavelength. What would be the minimum uncertainty of its momentum in this case? What part of the momentum does this uncertainty represent? Consider motion in one dimension.

  • Hint

    De Broglie wavelength of the particle is inversely proportional to its momentum. How exactly?

  • Solution

    We know that we can express the uncertainty of the location Δx as

    \[\Delta x=\frac{h}{p} .\]

    The location’s uncertainty Δx and uncertainty of momentum Δp are tied together through the Heisenberg principle of uncertainty:

    \[\Delta x\,\cdot\,\Delta p\ge\frac{\hbar}{2} .\]

    From that we can derive

    \[\Delta p \ge \frac{\hbar}{\,2\Delta x}=\frac{\hbar p}{2h}=\frac{\hbar p}{4\pi \hbar}=\frac{p}{4\pi}\,\dot{=}\, 0.08\,p .\]
  • Answer

    The smallest possible uncertainty of the particle’s momentum is approximately 8 % of its value.

Difficulty level: Level 3 – Advanced upper secondary level
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