We have found that de Broglie wavelength for a classical object (say an moving automobile or a ball) has very very tiny wavelength due to its bigger mass(as compared to the sub-atomic particles). So that kind of short wavelength is very much ridiculous & thus we neglect the wave nature for classical bodies. But why should this short wavelength be neglected? It has a very very high frequency! So why should we neglect that frequency?! By considering only the wavelength we reject the wave nature. But why can't we think about its frequency?
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1I find it heuristic to argue with de Broglie wavelength. – SRS Nov 21 '16 at 12:37
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1Diffractive effects (that reveal the "waviness" of something), only occur when you are considering apertures that are on the same order as the wavelength of the wave you are passing through them. The fact that classical objects have such a – Thomas Russell Nov 21 '16 at 13:15
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1Related, and possibly a duplicate: Validity of naively computing the de Broglie wavelength of a macroscopic object – John Rennie Nov 21 '16 at 17:02
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Just adding some numbers for completion, oblivion of the unrealistic scenario. For a spherical $1000$ Kg car modeled as a single De Broglie sine wave $\lambda = \frac{h}{m v}$ traveling in vacuum at speed $\lambda f =v =100$ Km/h then $f = \frac{m v^2}{h} \approx 10^{39}$ Hz – Rol Nov 30 '21 at 20:33
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The way you think about the frequency is like the "frequency of photons" which we assume to have an energy equal to $E=h\nu$ ($\nu=$ frequency).
For the particles the wavelength is related to the momentum by the relation $\lambda=\frac{h}{p}$. But it is not a photon to write for it $\lambda\nu=c$ ($c=$ speed of light).
However if you are insisting on having a relation between $\lambda$ and $\nu$, then you would get the relation $E=\frac{p^2}{2m}$ for the energy of the classical mass rewritten as a function of $\nu$.
Thomas Fritsch
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