where is the factorial value 1/2 of the exponent value in 10^500 #4.13 Correcting Math
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From: Owen Jacobson on 25 Jul 2010 12:48 On 2010-07-24 02:29:40 -0400, Archimedes Plutonium said: > Transfer Principle wrote: > >> solve 2x = log(x!) for x >> >> and then clicked on "use the base 10 logarithm." Though Wolfram >> returns the trivial value x = 0, there is a graph, and one can >> roll the mouse over the non-trivial solution: >> >> (268.087, 536.175) >> >> Thus, we find that 268! is approximately 10^536. >> > > 268! = 10^536 > Thanks, will store that for future use. > > I don't think there is a nucleus with 268 nucleons If you don't care for stability, you can fit rather more 268 nucleons into a nucleus. 285a Cn (element 112) has a half life around half a minute, which is just long enough to experiment with if you're quick (and behind a heavy radiation shield). 294 Uno (element 118) has a half life under 1 millisecond, which is still quite a while on the scale of nuclear reactions. Isotopes with more than 268 nucleons appear starting at dubnium (element 105). -o |