5 ms·
Extrapolating this, wouldn't that mean we could possibly get as much as 6.02e23 / (12 + C) bytes per mol of storage material? Where C is the number of atoms re
by sukuriant 15y ago
Extrapolating this, wouldn't that mean we could possibly get as much as 6.02e23 / (12 + C) bytes per mol of storage material?
Where C is the number of atoms required to communicate with that segment of memory. It might be
(6.02e23 / 12) + C, even!
That's a terrifying 5,569,618.84 petabytes for what could amount to a few hundred grams of materials (assuming it's a compound, and not just iron (at approx 56 grams a mol)).
Perhaps even many thousand petabytes in a storage space the size of a paperclip.
The potential is incredible; and it makes me worry: what will developers waste that space on next?
- wcoenen 15y agoThe article is about arranging atoms on a surface with a scanning tunnelling microscope. You you don't seem to be counting this required surface it in your "storage material". And that's still ignoring all the machinery needed to actually arrange/detect the atoms.
- sukuriant 15y agoThe reading material was considered with the C at the end of the calculation; but even still, if it takes 100 times as much equipment to read and write as it does to store, those are still ridiculous numbers. That was my point.