Calculate the bonding energy of two ions

• Engineering
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

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

Jaccobtw
Staff Emeritus
Homework Helper
Gold Member
$$E_N = \frac{-A}{r} + \frac{B}{r^{n}}$$
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.

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