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Homework Help: Factorial of avogadro number

  1. Nov 21, 2013 #1
    can anyone explain to me how to find the factorial of avogadro number ? what is its value ?
  2. jcsd
  3. Nov 21, 2013 #2


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    It's such a large number that you might be able to justify using Stirling's approximation.
  4. Nov 21, 2013 #3


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    Staff: Mentor

    Google for Stirling's approximation.
  5. Nov 21, 2013 #4


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    It's Huuuuuuuuuuuuuuuuuuuuge!
  6. Nov 22, 2013 #5
    I can tell you that the number of base 10 digits is > 6.02x10^23. You can't calculate this. Recently I have been coding my own big number c++ library. I often test it on 1,000,000!. It takes 5 minutes, and produces somewhere around 5 million digits. I doubt you are asked to calculate the exact value of 6.02E23!. If you want an approximation use Stirling approx. I use it in my e-calculator to find how many terms I need to sum to get the the desired accuracy.

    ln(n!) = n(ln(n) - 1) Note this is only a good approximation.
    This therefore means that n! = en(ln(n) - 1)) (approximately)
  7. Nov 29, 2013 #6


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    http://www.wolframalpha.com/ gives

    assumes usual metric value of
    6.022141×10^23 mol^(-1) (reciprocal moles)
    Last edited: Nov 29, 2013
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