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