Collection of Solved Problems in Physics

Particle as a wave

Task number: 4404

• Hint

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

• Solution of hint

  De Broglie wavelength of the particle with momentum p is equal to $$\lambda=\frac{h}{p} .$$

• 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.