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December 20, 2014

December 20, 2014

Posted by **Beth** on Friday, October 10, 2008 at 1:32pm.

n=(2J+1)exp(-(BJ(J+1))/(KT))

where B and K are constants

- Maths -
**drwls**, Friday, October 10, 2008 at 5:41pmHere's how you would do it for T.

ln n = ln (2J+1) -BJ(J+1)/kT

ln [n/(2J+1)] = -BJ(J+1)/kT

kT = -BJ(J+1)/ln[n/(2J+1)]

Now divide both sides by k

Work similarly to solve for J

- Maths -
**Beth**, Friday, October 10, 2008 at 6:17pmI understand how you do it for T, but since there is more than one J, I don't understand how you make J the subject.

- Maths -

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