The weight of brilliance

My Fermi estimate for the amount of data comprising is 215 GB. Suppose that all of this data is stored on SSD flash drives, which currently require about 10410^4 electrons per 1 bit, and that this data is information rich (Hint: a hard drive of all 1s or all 0s contains zero information). Under these assumptions, what is the mass (in kilograms) of the data that makes up


  • Vol(DB)=15 GB content+200 GB activity=215 GB data\text{Vol(DB)} = 15 \text{ GB content} + 200 \text{ GB activity} = 215 \text{ GB data}
  • 8 bits=1 byte8\text{ bits} = 1\text{ byte}
  • 1,073,741,824 bytes=1 gigabyte1,073,741,824\text{ bytes} = 1\text{ gigabyte}

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