Net Potential Energy between two Adjacent Ions

[Saladsamurai]

Homework Statement

The net energy is given by:

E_N=-\frac{A}{r}+\frac{B}{r^n}

where A, B , and n are constants and r is the interionic separation. Calculate E₀ in terms of A, B, and n by the following procedure:

1. find dE_N/dr
2. set this expression equal to zero and solve for r=r_o
3. substitute r_o back into the original equation

The Attempt at a Solution

Okay, this is more or less an algebra problem that I am stuck on:

E_N=-\frac{A}{r}+\frac{B}{r^n}

=-Ar^{-1}+Br^{-n}

\Rightarrow \frac{dE_N}{dr}=Ar^{-2}-nBr^{-n-1}

0=Ar^{-2}-nBr^{-n-1}

\Rightarrow 0=\frac{A}{r^2}-\frac{nB}{r^{n+1}}

\Rightarrow 0=\frac{Ar^{n+1}-nBr^2}{r^2*r^{n+1}}

\Rightarrow 0=Ar^{n+1}-nBr^2

Here is where my brain melted. Any blatant errors and/or hints?

Hmmm delicious hints

 

[Thaakisfox]

factor it out:

0=r^2(Ar^{n-1}-nB)

r_0=0 is nonsense, so we have:

r_0^{n-1}=\frac{nB}{A}

Now plug this back, and you are done.. :D

 

[Saladsamurai]

Egads man! That was easy. Nice catch Thaakisfox

Except that r_0^{n-1}=\frac{nB}{A} is not what I plug back in; I still have to solve explicitly for r which means I need to take the (n-1)th root of nB/A
right?

 

[Saladsamurai]

What am I still missing here?

If:
r=(\frac{nB}{A})^{\frac{1}{n-1}}

then:

E_0=-\frac{A}{(\frac{nB}{A})^{\frac{1}{n-1}}}+\frac{B}{(\frac{nB}{A})^{\frac{n}{n-1}}}

which is just silly.

 

[Saladsamurai]

I am thinking that this just does not clean up any better than this; i am not sure why I assumed that it would