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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

Calculate force (F) as a function of interionic distance (r) using the given EN equation.

We can rewrite the given equation for the net bonding energy as follows. $$ E_{N}(r)=-\frac{A}{r}+\frac{B}{r^n}. $$ The force (F) between the two ions is equal to the derivative of the net bonding energy (EN) with respect to the interionic distance (r). $$ F(r) = \frac{dE_{N}(r)}{dr}. $$ Taking the derivative of the EN equation with respect to r, we obtain: $$ F(r) = \frac{d}{dr}\left(-\frac{A}{r} + \frac{B}{r^n}\right). $$ This simplifies to: $$ F(r) = \frac{A}{r^2} - \frac{nB}{r^{n+1}}. $$

02

Take the derivative of F(r) with respect to r.

Now, let's find the derivative of the force function F(r) with respect to r: $$ \frac{dF(r)}{dr} = \frac{d}{dr}\left(\frac{A}{r^2} - \frac{nB}{r^{n+1}}\right). $$ This can be simplified to: $$ \frac{dF(r)}{dr} = \frac{-2A}{r^3} + \frac{n(n+1)B}{r^{n+2}}. $$

03

Find the equilibrium separation (r0).

At equilibrium separation (r0), the force value will be at the minimum of the EN-versus-r curve. This means that the derivative of EN with respect to r will be equal to zero. Using the force equation F(r) we found earlier, we have: $$ 0 = \frac{A}{r^2} - \frac{nB}{r^{n+1}}. $$ Rearranging this equation and solving for r, we get: $$ \frac{A}{r^2} = \frac{nB}{r^{n+1}} \implies r_0 = \left(\frac{nB}{A}\right)^{\frac{1}{n-1}}. $$

04

Substitute expression for r0 into the relationship derived by taking dF/dr.

Now we will substitute the expression for r0 that we found in Step 3 into the equation we derived for dF/dr in Step 2: $$ \left(\frac{dF}{dr}\right)_{r_{0}} = \frac{-2A}{r_{0}^3} + \frac{n(n+1)B}{r_{0}^{n+2}}. $$ Now, we substitute the expression of equilibrium separation that we found in Step 3: $$ \left(\frac{dF}{dr}\right)_{r_{0}} = \frac{-2A}{\left(\frac{nB}{A}\right)^{\frac{3}{n-1}}} + \frac{n(n+1)B}{\left(\frac{nB}{A}\right)^{\frac{n+2}{n-1}}} $$ Since the modulus of elasticity E is proportional to dF/dr at r0, we can establish a dependence of E on A, B, and n as: $$ E \propto \frac{-2A}{\left(\frac{nB}{A}\right)^{\frac{3}{n-1}}} + \frac{n(n+1)B}{\left(\frac{nB}{A}\right)^{\frac{n+2}{n-1}}}. $$ This equation shows how the modulus of elasticity E depends on A, B, and n for the two-ion system.

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