Equipartition theorem and chlorine

[PeterPoPS]

[SOLVED] Equipartition theorem

Homework Statement

If the chlorine molecule at 290 K were to rotate at the angular frequency predicted by the equipartition theorem what would be the average centripetal force?
(The atmos of Cl are 2*10^-10 m apart and the mass of the chlorine atom 35.45 a.m.u)

(Correct answer is 1.6*10^-10 N)

Homework Equations

Kinetic energy
E_{kin} = \frac{mv^2}{2}

Centripetal force
F_c = \frac{mv^2}{r}

Boltzmann's constant
k_B = 1.3807*10^{-23} JK^{-1}

The Attempt at a Solution

Hi!

I tried to solve it this way.
I think of my Chlorine molecule as two points. One stationary and the other one circulating around it.
From the equipartition theorem i get that the average energy of the molecule is

E = \frac{7k_BT}{2}

I think that this is equal to the kinetic energy

E = \frac{7k_BT}{2} = \frac{mv^2}{2}

so

mv^2 = 7k_BT

Using this in the equation for the centriopetal force I get

F_c = \frac{7k_BT}{r} = 1.4014^{-10}N

which is not the right answer and I havn't used the mass at all in the problem.
What's wrong??

 

[Shooting Star]

For diatomic molecules, Eavg = 5/2 KbT.