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: $$E_{N}=-\frac{A}{r}+\frac{B}{r^{n}}\quad\quad\quad (6.25)$$ 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. since \(r_{0}\) corresponds to the value of \(r\) at the minimum of the \(E_{N^{-} \text {versus- } r}\) 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

Establish a relationship for the force \(F\) as a function of \(r\)

Given that \(F=\frac{dE_N}{dr}\), let's differentiate the equation for \(E_N\) with respect to \(r\).

02

Differentiate \(E_N\) with respect to \(r\)

$$ F = \frac{dE_N}{dr} = \frac{d(-\frac{A}{r}+\frac{B}{r^n})}{dr} $$ Applying the chain rule of differentiation, we get $$ F = \frac{d(-Ar^{-1}+Br^{-n})}{dr} = A \frac{d(r^{-1})}{dr} - Bn \frac{d(r^{-n})}{dr} $$ Now, we've calculations: $$ \frac{d(r^{-1})}{dr} = -r^{-2} \quad\quad \text{and} \quad\quad \frac{d(r^{-n})}{dr} = -nr^{-n-1} $$ Substitute these back into the expression for \(F\): $$ F = A(-r^{-2}) - Bn(-nr^{-n-1}) = \frac{A}{r^2} - \frac{Bn}{r^{n+1}} $$

03

Develop an expression for \(r_0\), the equilibrium separation

To find the equilibrium separation \(r_0\), we need to find the point at which the bonding energy \(E_N\) is at a minimum. This corresponds to the value of \(r\) at which \(\frac{dE_N}{dr} = 0\). $$ 0 = \frac{A}{r_0^2} - \frac{Bn}{r_0^{n+1}} $$ Rearranging this equation to solve for \(r_0\): $$ \frac{A}{r_0^2} = \frac{Bn}{r_0^{n+1}} \Rightarrow r_0^{n+1-2} = r_0^{n-1} = \frac{Bn}{A} $$

04

Substitute the expression for \(r_0\) into the relationship obtained by taking \(dF/dr\)

Substitute the expression for \(r_0\) in terms of \(A\), \(B\), and \(n\) back into the expression for \(F\): $$ F = \frac{A}{r^2} - \frac{Bn}{r^{n+1}} $$ To find the expression for the modulus of elasticity \(E\), we need to find the derivative \(dF/dr\) and evaluate it at the equilibrium separation \(r_0\). $$ E \propto \left(\frac{dF}{dr}\right)_{r_0} \propto \left(\frac{d(\frac{A}{r^2} - \frac{Bn}{r^{n+1}})}{dr}\right)_{r_0} $$ Differentiate the expression for \(F\) with respect to \(r\): $$ \frac{dF}{dr} = -\frac{2A}{r^3} + \frac{(n+1)Bn}{r^{n+2}} $$ Now, evaluate this expression at \(r_0\): $$ \left(\frac{dF}{dr}\right)_{r_0} = -\frac{2A}{r_0^3} + \frac{(n+1)Bn}{r_0^{n+2}} $$ Using the expression for \(r_0\) in terms of \(A\), \(B\), and \(n\), we substitute back: $$ \left(\frac{dF}{dr}\right)_{r_0} = -\frac{2A}{( \frac{Bn}{A})^{3/(n-1) }} + \frac{(n+1)Bn}{( \frac{Bn}{A})^{(n+2)/(n-1)}} $$ Thus, we have derived an expression for the dependence of the modulus of elasticity on the constants \(A\), \(B\), and \(n\): $$ E \propto \left(\frac{dF}{dr}\right)_{r_0} $$

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