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From Wikipedia:

Light has a certain probability of being scattered by a material. When photons are scattered, most of them are elastically scattered (Rayleigh scattering), such that the scattered photons have the same energy (frequency, wavelength and color) as the incident photons but different direction. Rayleigh scattering usually has an intensity in the range 0.1% to 0.01% relative to that of a radiation source. An even smaller fraction of the scattered photons (approximately 1 in 10 million) can be scattered inelastically, with the scattered photons having an energy different (usually lower) from those of the incident photons—these are Raman scattered photons.[1

But, ALL of the photons hitting a particular solid object are ultimately scattered or absorbed-and-re-emitted somewhere, correct?

I can't imagine a photon in the greater visible-light range being entirely absorbed and converted to motion (usually)...

Also, Rayleigh scattering is the same thing as diffuse reflection, right?

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That the number of photons is preserved is not a physical law. There are a few examples in which an optical photon is "lost":

Both, Rayleigh scattering and diffuse reflection are elastic processes, but Rayleigh scattering usually refers to scattering of light with small (sub-$\lambda$) particles or even single atoms, while diffuse reflection is an effect of macroscopic surfaces.
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