Mean thermal energy of a system with given potential energy

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

$V(r) = \frac 1{r^3} \left( A - Br^2 \right )$
At the bottom of the well, r is verry small.
So, $V(r) = \frac A{r^3}$

Assuming the validation of Equipartition of energy theorem, since the degrees of freedom is 1,
the particle's mean thermal energy is $\frac { k_B T} 2$.
Is this correct so far?

 

[DrClaude]

$V(r) = \frac 1{r^3} \left( A - Br^2 \right )$

Where does the 3 come from?

At the bottom of the well, r is verry small.
So, $V(r) = \frac A{r^3}$

Does that show that "the bottom of the well is approximately quadratic in r"?

 

[Pushoam]

Where does the 3 come from?

I have taken n= 3.

Does that show that "the bottom of the well is approximately quadratic in r"?

No, this is the problem. The question says to show it in quadratic form, but it is not so.

 

[kuruman]

Leave $n$ alone. Suppose you were to find at what value of $r=r_0$ the potential has a minimum. Consider placing a particle at that minimum and displacing it slightly. What kind of motion do you think it will undergo?