Question

The net potential energy between two adjacent ions, EN, may be represented by the sum of Equations 2.9 and 2.11; that is, (2.17) EN= A/r+B/r^n Calculate the bonding energy E0 in terms of the parameters A, B, and n using the following procedure:
1. Differentiate EN with respect to r, and then set the resulting expression equal to zero, because the curve of EN versus r is a minimum at E0.
2. Solve for r in terms of A, B, and n, which yields r0, the equilibrium interionic spacing.
3. Determine the expression for E0 by substituting r0 into Equation 2.17.

Answer 1

Answer:

Eo = A/[-nB/A]^(1/n-1) + B/[-nB/A]^(n/n-1)

Explanation:

Step 1.

Taking derivative of the equation with respect to 'r' we get:

d/dr(EN) = - A/r² - nB/r^(n+1)

Setting this equation to zero:

Step 2.

Solving for r:

- A/r² - nB/r^(n+1) = 0

A/r² + nB/r^(n+1) = 0

Ar^(n+1) + nBr² = 0

Ar^(n+1) = - nBr²

[r^(n+1)]/r² = - nB/A

r^(n+1-2) = - nB/A

r^(n-1) = - nB/A

Taking power 1/(n-1) on both sides:

r = [-nB/A]^(1/n-1)

This is the value of ro:

ro = [-nB/A]^(1/n-1)

Step 3.

Substituting value of ro in eqn we get value of Eo

Eo = A/[-nB/A]^(1/n-1) + B/[-nB/A]^(n/n-1)