Calculate the bonding energy of two ions

[Jaccobtw]

Homework Statement

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

Calculate the bonding energy $E_0$ in terms of the parameters A, B, and n using
the following procedure:
1. Differentiate $E_N$ with respect to r, and then set the resulting expression equal
to zero, because the curve of $E_N$ versus r is a minimum at $E_0$.
2. Solve for r in terms of A, B, and n, which yields $r_0$, the equilibrium interionic
spacing.
3. Determine the expression for $E_0$ by substituting $r_0$ for r

Relevant Equations

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

1.) So first I differentiate and set it equal to 0 and get:
$$\frac{A}{r^2} -\frac{Bn}{r^{n-1}} = 0$$

2.) When solving for r, I'm not quite sure how to take away the exponent so I get up to the second to last step:

$$r^{n-3} = \frac{Bn}{A}$$

Would it be:

$$r = \sqrt[n-3]{\frac{Bn}{A}}$$

?

Am I doing this problem correctly?

Thank you

 

[DrClaude]

Correct. You can note the result as $r_0$, i.e.,
$$
r_0 = \left( \frac{Bn}{A} \right)^{1/(n-3)}
$$

 

[SammyS]

$$E_N = \frac{-A}{r} + \frac{B}{r^{n}}$$
1.) So first I differentiate and set it equal to 0 and get:
$$\frac{A}{r^2} -\frac{Bn}{r^{n-1}} = 0$$
2.) When solving for r, I'm not quite sure how to take away the exponent so I get up to the second to last step:
$$r^{n-3} = \frac{Bn}{A}$$
Would it be:
$$r = \sqrt[n-3]{\frac{Bn}{A}}\ \ \ ?$$
Am I doing this problem correctly?

Thank you

You made an error in taking the derivative.

Writing $E_N$ as $\displaystyle \quad \quad E_N= -A\,r^{-1} + B\,r^{-n} \quad$ may help.

Then $\displaystyle \quad \quad \dfrac{E_N}{dr}= A\,r^{-2} - B\,r^{-n-1} = A\,r^{-2} - B\,r^{-(n+1)}$

$\displaystyle \quad \quad \quad \quad \quad \quad = \dfrac{A}{r^2} - \dfrac{nB}{r^{(n+1)}} \quad$