 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 can carry at most six bits in every encoding: accessible modes, energy, aperture, measurements and success probabilities define the channel.