= Solution
To distinguish $2^b$ equal transverse bins across width $w$, $\boxed{d=w/2^b}$. Six bits gives one-nanometre pixels; ten gives 0.0625 nm, below the <helium> wavelength. In the stipulated fixed-energy freely propagating beam, the latter entails transverse diffraction <momenta> comparable to or exceeding the available total <momentum> and lies outside the paraxial aperture model. Six-bit localization is less demanding, although fine detector resolution and losses remain practical constraints. This is not a universal theorem that an <atom (Physics)>[atom] can carry at most six bits in every encoding: accessible modes, <energy>, aperture, measurements and success probabilities define the channel.
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