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 \(6.31\) 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,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 2.10b), 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 \(d F / d r\).

Short Answer

Expert verified
The expression for the dependence of the modulus of elasticity, E, on the constants A, B, and n, as well as the equilibrium separation r₀ is: $$ E \propto -2A\frac{1}{r_0^3} +(n+1)nB\frac{1}{r_0^{n+2}} $$

Step by step solution

01

Find the force F as a function of r

First, we need to find the force \(F\) as a function of \(r\) by taking the derivative of the net bonding energy \(E_{N}\) with respect to \(r\). From the given equation, we have: $$ E_N=-\frac{A}{r}+\frac{B}{r^n} $$ The force \(F\) is given by: $$ F=\frac{dE_N}{dr} $$ Now, let's find the derivative of \(E_N\) with respect to \(r\).
02

Calculate the derivative dF/dr

We have: $$ F=\frac{dE_N}{dr} $$ Taking the derivative of \(E_N\) with respect to \(r\), we get: $$ F = A\frac{d(-\frac{1}{r})}{dr} + B\frac{d(\frac{1}{r^n})}{dr} $$ $$ F = A\frac{1}{r^2} - nB\frac{1}{r^{n+1}} $$
03

Calculate the equilibrium separation r₀

At the equilibrium separation \(r=r_0\), the force F is zero, so: $$ F = A\frac{1}{r_0^2} - nB\frac{1}{r_0^{n+1}} = 0 $$ Rearranging the terms: $$ \frac{A}{r_0^2} = nB\frac{1}{r_0^{n+1}} $$ Hence, the equilibrium separation \(r_0\) is given by: $$ r_0^n=n\frac{B}{A}r_0 $$
04

Substitute the expression for r₀ into the relationship dF/dr

Now, let's substitute the expression for \(r_0\) found in Step 3, into the relationship of \(F\) obtained in Step 2: $$ E\propto \frac{dF}{dr} \mid_{r=r_0} $$ So, $$ E \propto \frac{d}{dr}(A\frac{1}{r^2} - nB\frac{1}{r^{n+1}}) \mid_{r=r_0} $$ Taking the derivative: $$ E \propto -2A\frac{1}{r^3} +(n+1)nB\frac{1}{r^{n+2}} \mid_{r=r_0} $$ Plug in the value of \(r=r_0\): $$ E \propto -2A\frac{1}{r_0^3} +(n+1)nB\frac{1}{r_0^{n+2}} $$ We have now derived an expression for the dependence of the modulus of elasticity \(E\) on the constants \(A\), \(B\), and \(n\), as well as the equilibrium separation \(r_0\). The expression is: $$ E \propto -2A\frac{1}{r_0^3} +(n+1)nB\frac{1}{r_0^{n+2}} $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Interionic Force-Separation Curve
The interionic force-separation curve is a graphical representation of the force between ions as a function of their distance apart. As two ions approach each other, the force between them changes. At large separations, the force is negligible; as they get closer, the attractive force increases until it reaches a maximum. Any closer, and the ions begin to repel each other due to electron cloud overlap.

To understand the concept better, think of two magnets. When they are far apart, they do not influence each other, but as you bring them closer, they suddenly snap together. Now, if you tried compressing them further, it would be difficult because the same poles would repel each other. This is similar to how ions behave, and the curve that depicts this relationship is crucial for understanding a material's properties.

In the exercise, to establish a relationship for the force (\f\(F\f\)) as a function of interionic distance (\f\(r\f\)), the derivative of the net bonding energy (\f\(E_N\f\)) with respect to \f\(r\f\) is taken, which is the mathematical way to determine the slope of the curve at any point. This slope reflects how rapidly the force changes with distance – a steep slope indicates a strong dependency of force on distance, which is related to a material's stiffness, or modulus of elasticity.
Bonding Energy in Materials
Bonding energy in materials is the energy that holds atoms together within a crystal structure. It is an essential concept in understanding material stability and properties. It is the sum of all the attractive and repulsive forces between the ions. When these forces are balanced, the material is at a stable state referred to as the equilibrium interionic separation.

The bonding energy equation given in the exercise (\(E_N=-\frac{A}{r}+\frac{B}{r^n}\)) expresses this energy as a function of the distance between ions, '\f\(r\f\)', where '\f\(A\f\)' and '\f\(B\f\)' are constants related to the strength of attractive and repulsive forces, respectively, and '\f\(n\f\)' is a constant that depends on the nature of the interaction.

Minimizing Energy for Stability

In any system, the natural tendency is to minimize energy to achieve stability. This is why the ions arrange themselves at a distance where the bonding energy is at its lowest—a point which we refer to as equilibrium. Understanding how to calculate and manipulate this energy is crucial for developing materials with specific mechanical properties, such as tensile strength or elasticity.
Equilibrium Interionic Separation
Equilibrium interionic separation is the point at which the force between adjacent ions in a crystal structure is at a minimum, reflecting a stable arrangement of atoms within the material. It is denoted by '\f\(r_0\f\)' in the exercise's equations.

The conditions of equilibrium can be determined mathematically by setting the first derivative of the net bonding energy with respect to '\f\(r\f\)' to zero and solving for '\f\(r\f\)'. This derivative represents the force, and when it equals zero, it signifies there is no net force acting on the ions – they are in a 'comfortable' distance where attractions meet repulsions perfectly.

Finding the Sweet Spot

The equilibrium interionic separation is like the 'sweet spot' for atomic arrangement. Each type of material, depending on its atomic properties, has a characteristic '\f\(r_0\f\)' where its structure is most stable. In practical applications, knowing the '\f\(r_0\f\)' helps engineers and scientists predict material behavior under various conditions, like temperature changes or mechanical stress. This fundamental concept links the microscopic atomic arrangements of a material to its macroscopic mechanical properties, such as the modulus of elasticity, which plays a pivotal role in designing and utilizing materials across all areas of engineering and science.

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Most popular questions from this chapter

List four major differences between deformation by twinning and deformation by slip relative to mechanism, conditions of occurrence, and final result.

For a brass alloy, the stress at which plastic deformation begins is \(345 \mathrm{MPa}(50,000 \mathrm{psi})\), and the modulus of elasticity is \(103 \mathrm{GPa}\left(15.0 \times 10^{6} \mathrm{psi}\right)\) (a) What is the maximum load that can be applied to a specimen with a cross- sectional area of \(130 \mathrm{~mm}^{2}\left(0.2 \mathrm{in} .{ }^{2}\right)\) without plastic deformation? (b) If the original specimen length is \(76 \mathrm{~mm}(3.0\) in.), what is the maximum length to which it can] be stretched without causing plastic deformation?

A cylindrical rod of steel \(\left(E=207 \mathrm{GPa}, 30 \times 10^{\circ}\right.\) psi) having a yield strength of \(310 \mathrm{MPa}(45,000\) psi) is to be subjected to a load of \(11,100 \mathrm{~N}\) (2500 \(\left.\mathrm{Ib}_{i}\right)\). If the length of the rod is \(500 \mathrm{~mm}(20.0 \mathrm{in}\).), what must be the diameter to allow an elongation of \(0.38 \mathrm{~mm}(0.015\) in.)?

The lower yield point for an iron that has an († average grain diameter of \(1 \times 10^{-2} \mathrm{~mm}\) is 230 MPa (33,000 psi). At a grain diameter of \(6 \times 10^{-3}\) \(\mathrm{mm}\), the yield point increases to \(275 \mathrm{MPa}(40,000\) psi). At what grain diameter will the lower yield point be \(310 \mathrm{MPa}(45,000 \mathrm{psi}) ?\)

(a) Compare planar densities (Section \(3.11\) and Problem 3.60) for the \((100),(110)\), and (111) planes for FCC. (b) Compare planar densities (Problem 3.61) for the (100), (110), and (111) planes for BCC.

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