Thursday

April 2, 2015

April 2, 2015

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 -

**Answer this Question**

**Related Questions**

Math - Rearrange for J; n=(2J+1)exp(-(BJ(J+1))/(KT)) where T, B and K are ...

Maths - Transform the following relationships/funtions into linear form and ...

Maths - P=(2J+1)exp(-(BJ(J+1))/kT) where B, k and T are constants a. Show that ...

Calculus II - What is sin(i) -- (sin of the imaginary number, i) What is cos(i...

Math/Calculus - How would I evaluate the following integral by using integration...

maths-complex numbers - by using the substitution w = z^3, find all the ...

(expi)^i - This question is not well defined. By 'expi' do you mean exp(i)? In ...

Physics/Math - A can of sardines is made to move along an x axis from x1 = 0.2 m...

calc: arc length - find the exact length of this curve: y = ( x^3/6 ) + ( 1/2x...

(expi)^i - can someone tell me how can we calculate (expi)^i. there is one ...