alleryThe de Broglie wavelength of a particle with momentum p is:
A$\lambda = h/p$ (the de Broglie formula)
B$\lambda = h \cdot p$ (incorrect product)
C$\lambda = p/h$ (incorrect ratio)
D$\lambda = h \cdot c$ (constant)
Answer & Solution
Correct answer: A. $\lambda = h/p$ (the de Broglie formula)
de Broglie: λ = h/p = h/(mv). h = Planck's constant.
Significant for tiny particles (electrons)
negligible for macroscopic.
Confirmed by Davisson-Germer electron diffraction (1927).
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