In Section \(2.6\) it was noted that the net bonding energy \(E_{N}\) between two isolated positive and negative ions is a function of interionic distance \(r\) as follows: $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ where \(A, B\), and \(n\) are constants for the particular ion pair. Equation \(6.25\) is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity \(E\) is proportional to the slope of the interionic force-separation curve at the equilibrium interionic separation; that is, $$ E \propto\left(\frac{d F}{d r}\right)_{r_{0}} $$ Derive an expression for the dependence of the modulus of elasticity on these \(A, B\), and \(n\) parameters (for the two-ion system) using the following procedure: 1\. Establish a relationship for the force \(F\) as a function of \(r\), realizing that $$ F=\frac{d E_{N}}{d r} $$ 2\. Now take the derivative \(d F / d r\). 3\. Develop an expression for \(r_{0}\), the equilibrium separation. Because \(r_{0}\) corresponds to the value of \(r\) at the minimum of the \(E_{N}\)-versus-r curve (Figure \(2.8 b\) ), take the derivative \(d E_{N} / d r\), set it equal to zero, and solve for \(r\), which corresponds to \(r_{0}\). 4\. Finally, substitute this expression for \(r_{0}\) into the relationship obtained by taking \(d F / d r\).

Step by step solution

01

Relationship between Force and Distance

The first step is to determine the relationship between force (F) and distance (r) using the net bonding energy (E_N) equation. We are given: $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ To find F, we need to take the derivative of E_N with respect to r: $$ F=\frac{d E_{N}}{d r} $$

02

Taking the derivative dF/dr

To find F, we take the derivative of E_N with respect to r. Applying the derivative to the given equation, we get: $$ F = \frac{d}{dr}\left(-\frac{A}{r}+\frac{B}{r^n}\right) $$ Using the derivative rules, we derive the following expression for F: $$ F = \frac{A}{r^2}-\frac{nB}{r^{n+1}} $$

03

Finding the equilibrium separation (r0)

Now, we need to find r0, the equilibrium separation. For this, we will take the derivative of E_N with respect to r and set it equal to zero, as it corresponds to the minimum of the E_N versus r curve. So: $$ 0 = \frac{d E_N}{d r} = \frac{A}{r^2}-\frac{nB}{r^{n+1}} $$ We will solve this equation for r, which corresponds to r0. Multiplying both sides by r^{n+1}, we get: $$ 0 = A r^{n-1} - nB $$ Now, isolate r^{n-1} on one side: $$ r^{n-1}=\frac{nB}{A} $$ Finally, take the (n-1)th root on both sides to solve for r0: $$ r_{0}=\left(\frac{nB}{A}\right)^\frac{1}{n-1} $$

04

Substituting r0 into the derivative dF/dr

Lastly, we will substitute the obtained expression for r0 into the relationship obtained by taking dF/dr: $$ E\propto\left(\frac{dF}{dr}\right)_{r_0} $$ Now, let's take the derivative of F with respect to r: $$ \frac{dF}{dr} = -\frac{2A}{r^3}+\frac{(n+1)nB}{r^{n+2}} $$ Now, substitute r0 in this expression: $$ E\propto -\frac{2A}{r_{0}^3}+\frac{(n+1)nB}{r_{0}^{n+2}} $$ Replace r0 with its expression from previous step: $$ E \propto -\frac{2A}{\left(\frac{nB}{A}\right)^{\frac{3}{n-1}}}+\frac{(n+1)nB}{\left(\frac{nB}{A}\right)^{\frac{n+2}{n-1}}} $$ This expression shows the dependence of modulus of elasticity (E) on the constants A, B, and n for a two-ion system.

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