Finding the Equilibrium Temperature

[vertices]

Homework Statement

following on from my thread in the Nuclear physics subforum, I am trying to work out the equilibrium temperature of three isotopes (Neon) in a state of equilibrium established by thermal contact with a bath of neutrons at a temperature T.

Homework Equations

$$\frac{[20Ne][n]}{[21Ne]}=exp(\frac{B(21)}{kT})$$

$$\frac{[21Ne][n]}{[22Ne]}=exp(\frac{B(22)}{kT})$$

where for example,
B(21Ne)=[Mass(20Ne) + Mass(neutron) - Mass(21Ne)]c$$^{2}$$
square brackets [..] represent concentrations

This is the relavent data for Neon:

A Ab mass (amu)
20 90.92 19.9924
21 0.26 20.9939
22 8.82 21.9914

The Attempt at a Solution

Using the equations to find T gives:

$$T=\frac{B(21Ne)-B(22Ne)}{kln\frac{[20Ne][22Ne]}{[21Ne]^2}}$$ (*)

Now, B(21Ne)-B(22Ne)
={[Mass(20Ne)+Mass(neutron)-Mass(21Ne)]}-{[Mass(21Ne)+Mass(neutron)-Mass(22Ne)]
=Mass(20Ne) + Mass(22Ne) - 2*Mass(21Ne)

which is NEGATIVE (for neon, as well as for other isotope triplets)!

Because the denominator of (*) is positive, the equilbrium temperature is NEGATIVE! Weisacker who calculated this temperature worked it out to be 2.9x10^9K. What am I doing wrong? (I work it out to be -4.6x10^9)

 

[Lojzek]

I don't see any data about the number of each izotopes. Maybe you used higher number for higher energy state?