The net potential energy between two adjacent ions, \(E_{N}\), may be represented by the sum of Equations \(2.8\) and \(2.9\); that is, $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ Calculate the bonding energy \(E_{0}\) in terms of the parameters \(A, B\), and \(n\) using the following procedure: 1\. Differentiate \(E_{N}\) with respect to \(r\), and then set the resulting expression equal to zero, because the curve of \(E_{N}\) versus \(r\) is a minimum at \(E_{0 \text { - }}\) 2\. Solve for \(r\) in terms of \(A, B\), and \(n\), which yields \(r_{0}\), the equilibrium interionic spacing. 3\. Determine the expression for \(E_{0}\) by substituting \(r_{0}\) into Equation \(2.11\).

Short Answer

Expert verified
Question: Determine the bonding energy \(E_0\) in terms of the parameters \(A, B\), and \(n\). Answer: \(E_0 = -A\left(\frac{A}{nB}\right)^{\frac{1}{n-1}} + B\left(\frac{A}{nB}\right)^{\frac{n}{n-1}}\)

Step by step solution

01

Differentiate \(E_N\) with respect to \(r\) and set it to zero

First, we need to find the derivative of \(E_N\) with respect to \(r\): $$ \frac{dE_N}{dr} = \frac{d}{dr} \left(-\frac{A}{r} + \frac{B}{r^n}\right) $$ Now, by differentiating each term with respect to \(r\) and then setting the expression equal to zero, we obtain: $$ \frac{dE_N}{dr} = \frac{A}{r^2} - \frac{nB}{r^{n+1}} = 0 $$
02

Solve for \(r_0\) in terms of \(A, B\), and \(n\)

Next, we need to find \(r_0\), the equilibrium interionic spacing, by solving the above equation for \(r\). Let's rearrange the equation first: $$ \frac{A}{r^2} = \frac{nB}{r^{n+1}} $$ Now, solve for \(r\): $$ r_0^{n-1} = \frac{nB}{A} $$ So we get: $$ r_0 = \left(\frac{nB}{A}\right)^{\frac{1}{n-1}} $$
03

Determine the expression for \(E_0\) by substituting \(r_0\) into Equation \(2.11\)

Finally, we can substitute the value of \(r_0\) into the equation of \(E_N\) to find the bonding energy \(E_0\): $$ E_0 = -\frac{A}{r_0} + \frac{B}{r_{0}^n} $$ Substitute the value of \(r_0\) found in Step 2: $$ E_0 = -A\left(\frac{A}{nB}\right)^{\frac{1}{n-1}} + B\left(\frac{A}{nB}\right)^{\frac{n}{n-1}} $$ Now, we have the bonding energy \(E_0\) in terms of the parameters \(A, B\), and \(n\).

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Key Concepts

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

Interionic Potential Energy
Understanding the concept of interionic potential energy is fundamental in the field of materials science and physics. Interionic potential energy refers to the energy that is associated with the electrostatic interactions between ions in a crystal lattice. In simple terms, it is the energy due to the attraction or repulsion between charged particles.

From a practical standpoint, the equation \( E_{N} = -\frac{A}{r} + \frac{B}{r^{n}} \) allows us to model how the energy varies with the distance \( r \) between two ions. The first term (negative) represents the attractive forces, while the second term (positive) accounts for the repulsive forces that come into play at shorter distances. The constants \( A \) and \( B \) characterize the strength of these forces, and \( n \) is an exponent that depends on the specific interaction.

It is essential to recognize that the potential energy is at its minimum when the ions are at an equilibrium spacing, which brings us to our next concept: equilibrium interionic spacing.
Equilibrium Interionic Spacing
The equilibrium interionic spacing, represented as \( r_{0} \) in materials science, is the distance between two ions at which the interionic potential energy is minimized. This spacing is crucial because it reflects the most stable arrangement of ions in a lattice - essentially, it's the distance at which the material would prefer to stay under ambient conditions.

To find this optimized point mathematically, we differentiate the interionic potential energy with respect to \( r \) and then set this derivative equal to zero to find the value of \( r \) that minimizes the energy (Step 1 and Step 2 in the solution). This process is effectively looking for a 'sweet spot' where the forces of attraction and repulsion between ions are perfectly balanced.

Through this differentiation, we arrive at the equation: \( r_0 = \left(\frac{nB}{A}\right)^{\frac{1}{n-1}} \). This equation provides us with a direct way to calculate the spacing using the constants from the potential energy equation, assuming we have determined them experimentally or through other theoretical models.
Differentiation in Materials Science
Differentiation is a powerful mathematical tool extensively used in materials science to analyze and solve problems involving variable relationships. In the context of bonding energy calculations, differentiation helps us determine the conditions for equilibrium and stability within a material's structure.

By differentiating the interionic potential energy with respect to the interionic distance \( r \) and equating it to zero, we can find the point of equilibrium spacing (Step 1). This is a practical application of differentiation, which transforms an energy-distance relationship into valuable information about the material's properties, such as its bonding energy at equilibrium (Step 2 and Step 3).

The importance of such differentiation is not merely academic; it lays the foundation for predicting material behavior under different conditions, guiding the design of new materials and enhancing the understanding of existing materials. The ability to predict the equilibrium state means that scientists and engineers can estimate many other properties of the material, such as its strength, thermal expansion, and electronic properties.

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Most popular questions from this chapter

6 Allowed values for the quantum numbers of electrons are as follows: $$ \begin{aligned} n &=1,2,3, \ldots \\ l &=0,1,2,3, \ldots, n-1 \\ m_{l} &=0, \pm 1, \pm 2, \pm 3, \ldots, \pm l \\ m_{s} &=\pm \frac{1}{2} \end{aligned} $$ The relationships between \(n\) and the shell designations are noted in Table 2.1. Relative to the subshells, \(l=0\) corresponds to an \(s\) subshell \(l=1\) corresponds to a \(p\) subshell \(l=2\) corresponds to a \(d\) subshell \(l=3\) corresponds to an \(f\) subshell For the \(K\) shell, the four quantum numbers for each of the two electrons in the \(1 s\) state, in the order of \(n l m_{i} m_{s}\), are \(100 \frac{1}{2}\) and \(100\left(-\frac{1}{2}\right)\) Write the four quantum numbers for all of the electrons in the \(L\) and \(M\) shells, and note which correspond to the \(s, p\), and \(d\) subshells.

Relative to electrons and electron states, what does each of the four quantum numbers specify?

Give the electron configurations for the following ions: \(\mathrm{Fe}^{2+}, \mathrm{Al}^{3+}, \mathrm{Cu}^{+}, \mathrm{Ba}^{2+}, \mathrm{Br}^{-}\), and \(\mathrm{O}^{2-}\)

(a) How many grams are there in one amu of a material? (b) Mole, in the context of this book, is taken in units of gram-mole. On this basis, how many atoms are there in a pound-mole of a substance?

Chromium has four naturally occurring isotopes: \(4.34 \%\) of \({ }^{50} \mathrm{Cr}\), with an atomic weight of \(49.9460\) amu; \(83.79 \%\) of \({ }^{52} \mathrm{Cr}\), with an atomic weight of \(51.9405 \mathrm{amu} ; 9.50 \%\) of \({ }^{53} \mathrm{Cr}\), with an atomic weight of \(52.9407 \mathrm{amu} ;\) and \(2.37 \%\) of \({ }^{54} \mathrm{Cr}\), with an atomic weight of \(53.9389\) amu. On the basis of these data, confirm that the average atomic weight of \(\mathrm{Cr}\) is \(51.9963 \mathrm{amu}\)

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