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Evaluating an exponential of a really large negative number

  1. Jun 5, 2009 #1
    As a kid, I remember my father saying "there's a small chance that an electron in your body is on the moon" Well, today I decided to calculate the odds. Among the assumptions I made to make math easier.

    *Ground state wave function of Hydrogen
    *the moon is a cube of sides 2r. where r is the radius of the moon.
    *Ignore gravity

    Setting the origin at earth, you simply integrate over the volume of the moon in spherical coordinates.

    I don't have the result on me now, so I'm guessing at what I got. I think I eliminated some pi's and constants because it's an order of magnitude kind of situation.

    { (rm^2)e^(-ro/ao) } / (ro^2)

    rm - radius of the moon
    ro - radius of the moon's orbit
    ao - bohr radius

    My question, the term e^(-ro/ao) is an exponential of a huge negative number. The grapher on this mac makes things look pretty, but it can't crunch numbers. How would you evaluate this?
  2. jcsd
  3. Jun 5, 2009 #2


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    For numerical evaluation of large (negative or positive) powers of e, convert first to a power of 10, by multiplying the exponent by log10e. The rest is obvious.
  4. Jun 5, 2009 #3
    So I want to take both sides of the equation to the log10e power?
  5. Jun 5, 2009 #4


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    Well... uh, sort of. You'll want to use the identity
    [tex]e^{-r_0/a_0} = 10^{-(r_0/a_0)\log e}[/tex]
    So just figure out what [itex](-(r_0/a_0)\log e)[/itex] is, and then if it's, say, -1000000000 you'll have an answer like "10 to the -1000000000 power" or whatever. Since it's just an order-of-magnitude thing, that's good enough.
  6. Jun 6, 2009 #5
    Ok. I atleast see that that equation is balanced. Good enough for me.

    What a crazy chance diazona, My calculations end up with...

    p = 10^-(10^8) = 10^-10,000,000
  7. Jun 6, 2009 #6
    I'd like to see a cubic moon actually.
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