Momentum-wavelength Relation

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Particle momentum in terms of de Broglie wavelength λ?

RankGuru · Accepted Answer

1. **Fundamental Equation**: The de Broglie relation is $\lambda = \frac{h}{p}$.
2. **Rearrange for Momentum**: Cross-multiplying gives $p = \frac{h}{\lambda}$.
3. **Context**: This relation links a 'particle' property (linear momentum $p$) to a 'wave' property (wavelength $\lambda$) via Planck's constant $h$.
4. **Conclusion**: The momentum is inversely proportional to the associated wavelength.

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Momentum-wavelength Relation

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