Posted by Bach on Wednesday, April 28, 2010 at 1:23am.
Yes but I would have gone about it another way.
1312000/6.022 x 10^23 = 2.18 x 10^-18 J/atom by your calculations.
I would use
delta E = 2.180 x 10^-18 J (1/N1^2 -1/N2^2)
For N2 = infinity, 1/N^2 is zero and 1/N1^2 = 1; therefore, E = 2.180 x 10^18 joules/atom
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